Differential Equations 9.3 NCERT Solutions

1. Find the general solution of the following differential equation:

\displaystyle \frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}

Solution

Think First… The given differential equation is already in variable separable form since the right-hand side contains only x. Move dy to the left and dx to the right, then integrate both sides.

Given,

\displaystyle \frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}

Separating the variables, we get

\displaystyle dy=\frac{1-\cos x}{1+\cos x}\,dx

Integrating both sides,

\displaystyle \int dy=\int\frac{1-\cos x}{1+\cos x}\,dx

Using the identity

\displaystyle \frac{1-\cos x}{1+\cos x}=\tan^2\frac{x}{2}

we get

\displaystyle \int dy=\int\tan^2\frac{x}{2}\,dx

Now, using the identity

\displaystyle \tan^2\theta=\sec^2\theta-1

we obtain

\displaystyle \int dy=\int\left(\sec^2\frac{x}{2}-1\right)dx

Integrating each term separately,

\displaystyle y=\int\sec^2\frac{x}{2}\,dx-\int dx+C

Using the substitution

\displaystyle t=\frac{x}{2}\quad\Rightarrow\quad dx=2\,dt

we get

\displaystyle \int\sec^2\frac{x}{2}\,dx=2\int\sec^2t\,dt

\displaystyle =2\tan t=2\tan\frac{x}{2}

Also,

\displaystyle \int dx=x

Therefore,

\displaystyle y=2\tan\frac{x}{2}-x+C

Hence, the required general solution is

\boxed{\displaystyle y=2\tan\frac{x}{2}-x+C}

2. Find the general solution of the following differential equation:

\displaystyle \frac{dy}{dx}=\sqrt{4-y^2}\; (-2\lt y\lt 2)

Solution

Think First… The given differential equation is already in variable separable form since the right-hand side contains only y. Separate the variables, integrate both sides, and use the standard integral of \displaystyle \frac{1}{\sqrt{a^2-x^2}}.

Given,

\displaystyle \frac{dy}{dx}=\sqrt{4-y^2}

Separating the variables, we get

\displaystyle \frac{dy}{\sqrt{4-y^2}}=dx

Integrating both sides,

\displaystyle \int\frac{dy}{\sqrt{4-y^2}}=\int dx

Using the standard formula

\displaystyle \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\left(\frac{x}{a}\right)+C

with a=2, we get

\displaystyle \sin^{-1}\!\left(\frac{y}{2}\right)=x+C

Taking sine on both sides,

\displaystyle \frac{y}{2}=\sin(x+C)

Therefore,

\displaystyle y=2\sin(x+C)

Hence, the required general solution is

\boxed{\displaystyle y=2\sin(x+C)}

3. Find the general solution of the following differential equation:

\displaystyle \frac{dy}{dx}+y=1\qquad (y\ne 1)

Solution

Think First… Rewrite the equation so that the derivative is alone on one side. Then separate the variables by keeping all the y terms on the left and the x terms on the right before integrating.

Given, \displaystyle \frac{dy}{dx}+y=1

or \displaystyle \frac{dy}{dx}=1-y

Separating the variables, we get

\displaystyle \frac{dy}{1-y}=dx

Integrating both sides,

\displaystyle \int\frac{dy}{1-y}=\int dx     …(1)

Using the standard formula

\displaystyle \int\frac{dx}{a-x}=-\log|a-x|+C

with a=1, we get from (1)

\displaystyle -\log|1-y|=x+C

or \displaystyle \log|1-y|=-x+C

Using definition of logarithms, we get

\displaystyle |1-y|=e^{-x+C}\; (\because\ \log_e(y)=u\ \Rightarrow\ y=e^u,\ \text{where }u=-x+C)

Using the law of exponents,

\displaystyle |1-y|=e^{-x}\cdot e^C

Since \displaystyle e^C is an arbitrary positive constant, let \displaystyle C'=e^C. Then

\displaystyle |1-y|=C'e^{-x}

Absorbing the sign into the arbitrary constant, we get

\displaystyle 1-y=C'e^{-x}

or \displaystyle y=1-C'e^{-x}

Let \displaystyle A=-C' to match the NCERT answer. Then

\boxed{\displaystyle y=1+Ae^{-x}}

4. Find the general solution of the following differential equation:

\displaystyle \sec^2x\,\tan y\,dx+\sec^2y\,\tan x\,dy=0

Solution

Think First… Separate the variables by keeping all the y terms on one side and all the x terms on the other. Then use the standard integral of \displaystyle \frac{\sec^2x}{\tan x}.

Given,

\displaystyle \sec^2x\,\tan y\,dx+\sec^2y\,\tan x\,dy=0

Transposing the second term to the right side, we get

\displaystyle \sec^2x\,\tan y\,dx=-\sec^2y\,\tan x\,dy

Separating the variables,

\displaystyle \frac{\sec^2y}{\tan y}\,dy=-\frac{\sec^2x}{\tan x}\,dx

Integrating both sides,

\displaystyle \int\frac{\sec^2y}{\tan y}\,dy=-\int\frac{\sec^2x}{\tan x}\,dx     …(1)

Using the substitution on LHS of (1),

\displaystyle t=\tan y\quad\Rightarrow\quad dt=\sec^2y\,dy

we get

\displaystyle \int\frac{\sec^2y}{\tan y}\,dy=\int\frac{dt}{t}=\log|t|=\log|\tan y|

Similarly on RHS of (1), using

\displaystyle u=\tan x\quad\Rightarrow\quad du=\sec^2x\,dx

we obtain

\displaystyle \int\frac{\sec^2x}{\tan x}\,dx=\int\frac{du}{u}=\log|u|=\log|\tan x|

Therefore, equation (1) becomes

\displaystyle \log|\tan y|=-\log|\tan x|+\log C

or \displaystyle \log|\tan x|+\log|\tan y|=\log C

(Note that, we have taken \log C as the constant of integration)

Using the property

\displaystyle \log a+\log b=\log(ab)

We get, \displaystyle \log|\tan x\,\tan y|=\log C

Taking anti-logs on both sides, we get

\displaystyle \tan x\,\tan y=C

Hence, the required general solution is

\boxed{\displaystyle \tan x\,\tan y=C}

5. Find the general solution of the following differential equation:

\displaystyle (e^x+e^{-x})\,dy-(e^x-e^{-x})\,dx=0

Solution

Think First… Separate the variables first. Then notice that the numerator is the derivative of the denominator. This converts the integral into the standard form \displaystyle \int\frac{f'(x)}{f(x)}\,dx.

Given,

\displaystyle (e^x+e^{-x})\,dy-(e^x-e^{-x})\,dx=0

Transposing the second term to the right side, we get

\displaystyle (e^x+e^{-x})\,dy=(e^x-e^{-x})\,dx

Separating the variables,

\displaystyle dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx

Integrating both sides, we get

\displaystyle \int dy=\int\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx     …(1)

Let \displaystyle t=e^x+e^{-x}

Then, \displaystyle dt=(e^x-e^{-x})\,dx

Substituting on RHS of (1), we get

\displaystyle \int\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx=\int\frac{dt}{t}

Using the standard integral

\displaystyle \int\frac{dx}{x}=\log|x|+C

we obtain

\displaystyle \int\frac{dt}{t}=\log|t|     …(2)

Substituting the value of t in (2), equation (1) reduces to,

\displaystyle y=\log|e^x+e^{-x}|+C

Since \displaystyle e^x+e^{-x}\gt0 for all real values of x, the modulus sign is not required.

Hence, the required general solution is

\boxed{\displaystyle y=\log(e^x+e^{-x})+C}

6. Find the general solution of the following differential equation:

\displaystyle \frac{dy}{dx}=(1+x^2)(1+y^2)

Solution

Think First… This is a variable separable differential equation. Separate the variables first and then use the standard integral of \displaystyle \frac{1}{1+x^2}, whose integral is \displaystyle \tan^{-1}x.

Given,

\displaystyle \frac{dy}{dx}=(1+x^2)(1+y^2)

Separating the variables, we get

\displaystyle \frac{dy}{1+y^2}=(1+x^2)\,dx

Integrating both sides,

\displaystyle \int\frac{dy}{1+y^2}=\int(1+x^2)\,dx     …(1)

Using the standard integrals

\displaystyle \int\frac{dx}{1+x^2}=\tan^{-1}x+C

and

\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\ne-1

we obtain from equation (1),

\displaystyle \tan^{-1}y=x+\frac{x^3}{3}+C

Hence, the required general solution is

\boxed{\displaystyle \tan^{-1}y=x+\frac{x^3}{3}+C}

7. Find the general solution of the following differential equation:

\displaystyle y\log y\,dx-x\,dy=0

Solution

Think First… Separate the variables first. Then rewrite the left side as \displaystyle \frac{dy}{y\log y}. Use the substitution \displaystyle t=\log y, since the derivative of \displaystyle \log y is \displaystyle \frac{1}{y}.

Given, \displaystyle y\log y\,dx-x\,dy=0

Transposing the second term to the right side, we get

\displaystyle y\log y\,dx=x\,dy

Separating the variables,

\displaystyle \frac{dy}{y\log y}=\frac{dx}{x}

Integrating both sides,

\displaystyle \int\frac{dy}{y\log y}=\int\frac{dx}{x}     …(1)

Let \displaystyle t=\log y

Then, \displaystyle dt=\frac{dy}{y}

Substituting in (1), we get

\displaystyle \int\frac{dt}{t}=\int\frac{dx}{x}

On integrating, using the standard integral, we obtain

\displaystyle \log|t|=\log|x|+\log C

(Note that, we have taken \log C as the constant of integration just like in Q. 4 above)

Substituting the value of t,

\displaystyle \log|\log y|=\log|x|+\log C

Using the log property on RHS

\displaystyle \log a+\log b=\log(ab)

we obtain

\displaystyle \log|\log y|=\log|Cx|

Taking anti-logs on both sides, we obtain

\displaystyle \log y=Cx

Using definition of logarithms, we get

\displaystyle y=e^{Cx}

\displaystyle (\because\ \log_e y=u\ \Rightarrow\ y=e^u,\ \text{here }u=Cx)

Hence, the required general solution is

\boxed{\displaystyle y=e^{Cx}}

8. Find the general solution of the following differential equation:

\displaystyle x^5\frac{dy}{dx}=-y^5

Solution

Think First… This is a variable separable differential equation. Separate the variables first and then use the standard integral of \displaystyle x^n, where \displaystyle n\ne-1, on both sides.

Given, \displaystyle x^5\frac{dy}{dx}=-y^5

Separating the variables, we get

\displaystyle \frac{dy}{y^5}=-\frac{dx}{x^5}

Integrating both sides,

\displaystyle \int y^{-5}\,dy=-\int x^{-5}\,dx

Using the standard integral

\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\quad n\ne-1

we get

\displaystyle -\frac{1}{4y^4}=\frac{1}{4x^4}+C     …(1)

Multiplying both sides of equation (1) by -4, we obtain

\displaystyle \frac{1}{x^4}+\frac{1}{y^4}=C'   where \displaystyle C'=-4C

Hence, the required general solution is

\boxed{\displaystyle x^{-4}+y^{-4}=C'}

9. Find the general solution of the following differential equation:

\displaystyle \frac{dy}{dx}=\sin^{-1}x

Solution

Think First… Since the right side is a function of x only, the variables are already separated. Integrate both sides directly. To evaluate \displaystyle \int\sin^{-1}x\,dx, use the method of Integration by Parts.

Given,

\displaystyle \frac{dy}{dx}=\sin^{-1}x

Integrating both sides, we get

\displaystyle \int dy=\int\sin^{-1}x\,dx     …(1)

Using Integration by Parts on RHS, i.e.

\displaystyle \int u\,v\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx

Consider, \displaystyle \int \underbrace{\sin^{-1}x}_{\text{I}}\;\underbrace{1}_{\text{II}}\,dx

Here, \displaystyle u=\sin^{-1}x,\qquad v=1

Therefore,

\displaystyle \frac{du}{dx}=\frac1{\sqrt{1-x^2}},\qquad \int v\,dx=\int1\,dx=x

Substituting in the formula above, we get

\displaystyle \int\sin^{-1}x\,dx=x\sin^{-1}x-\int\frac{x}{\sqrt{1-x^2}}\,dx  …(2)

To evaluate the remaining integral in (2) above,

Let \displaystyle t=1-x^2

Then, \displaystyle dt=-2x\,dx

or \displaystyle x\,dx=-\frac12\,dt

Substituting in equation (2), we get

\displaystyle \int\sin^{-1}x\,dx=x\sin^{-1}x+\frac12\int t^{-1/2}\,dt

Using the standard integral

\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\ne-1

we get

\displaystyle \int\sin^{-1}x\,dx=x\sin^{-1}x+t^{1/2}

Substituting the value of t,

\displaystyle \int\sin^{-1}x\,dx=x\sin^{-1}x+\sqrt{1-x^2}

Therefore, equation (1) gives

\displaystyle y=x\sin^{-1}x+\sqrt{1-x^2}+C

Hence, the required general solution is

\boxed{\displaystyle y=x\sin^{-1}x+\sqrt{1-x^2}+C}

10. Find the general solution of the following differential equation:

\displaystyle e^x\tan y\,dx+(1-e^x)\sec^2y\,dy=0

Solution

Think First… Separate the variables first. Keep the right side in the form \displaystyle -\frac{e^x}{1-e^x}. Then use the substitution \displaystyle t=1-e^x to evaluate the integral in one simple step.

Given,

\displaystyle e^x\tan y\,dx+(1-e^x)\sec^2y\,dy=0

Separating the variables,

\displaystyle \frac{\sec^2y}{\tan y}\,dy=-\frac{e^x}{1-e^x}\,dx

Integrating both sides,

\displaystyle \int\frac{\sec^2y}{\tan y}\,dy=-\int\frac{e^x}{1-e^x}\,dx     …(1)

For the left side, let

\displaystyle t=\tan y

Then, \displaystyle dt=\sec^2y\,dy

Therefore, equation (1) reduces to

\displaystyle \log|\tan y|=-\int\frac{e^x}{1-e^x}\,dx    …(2)

To evaluate the integral on the right, let

\displaystyle t=1-e^x

Then, \displaystyle dt=-e^x\,dx

Substituting in equation (2), we get

\displaystyle \log|\tan y|=\int\frac{dt}{t}=\log|t|+\log C

Substituting back the value of t,

\displaystyle \log|\tan y|=\log|1-e^x|+\log C

Using the property \displaystyle \log a+\log b=\log(ab),

\displaystyle \log|\tan y|=\log|C(1-e^x)|

Therefore, taking anti-logs, we obtain

\displaystyle \tan y=C(1-e^x)

Hence, the required general solution is

\boxed{\displaystyle \tan y=C(1-e^x)}

11. Find the particular solution of the following differential equation satisfying the given condition:

\displaystyle (x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x;\:y(0)=1 \text{ i.e. }y=1\text{ when }x=0

Solution

Think First… First find the general solution by separating the variables. Then use the given condition \displaystyle y=1\text{ when }x=0 to determine the value of the arbitrary constant and hence obtain the particular solution.

Given,

\displaystyle (x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x

Since

\displaystyle x^3+x^2+x+1=(x+1)(x^2+1)

Separating the variables,

\displaystyle dy=\frac{x(2x+1)}{(x+1)(x^2+1)}\,dx     …(1)

Since the denominator contains one linear factor and one irreducible quadratic factor, we use partial fractions.

Let

\displaystyle \frac{x(2x+1)}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}

Multiplying both sides by \displaystyle (x+1)(x^2+1), we get

\displaystyle x(2x+1)=A(x^2+1)+(Bx+C)(x+1)

Expanding the right side,

\displaystyle 2x^2+x=A(x^2+1)+Bx^2+Bx+Cx+C

\displaystyle =(A+B)x^2+(B+C)x+(A+C)

Comparing the coefficients of like powers of x and constant terms, we get

\displaystyle A+B=2,\qquad B+C=1,\qquad A+C=0

From \displaystyle A+C=0,\text{ we get }C=-A.

Substituting this in \displaystyle B+C=1,

\displaystyle B-A=1     …(2)

Also, \displaystyle A+B=2     …(3)

Adding equations (2) and (3),

\displaystyle 2B=3 \Rightarrow B=\frac32

Substituting this value of B in equation (3),

We get, \displaystyle A=2-\frac32=\frac12

Also, \displaystyle C=-A=-\frac12

Hence,

\displaystyle A=\frac12,\qquad B=\frac32,\qquad C=-\frac12

Therefore,

\displaystyle \frac{x(2x+1)}{(x+1)(x^2+1)}=\frac1{2(x+1)}+\frac{3x-1}{2(x^2+1)}

Hence, integrating equation (1), we get

\displaystyle \int dy=\int\left(\frac1{2(x+1)}+\frac{3x}{2(x^2+1)}-\frac1{2(x^2+1)}\right)dx

Using the standard integrals

\displaystyle \int\frac{dx}{x}=\log|x|+C,\qquad \int\frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C,\qquad \int\frac{dx}{1+x^2}=\tan^{-1}x+C

we get

\displaystyle y=\frac12\log|x+1|+\frac34\log(x^2+1)-\frac12\tan^{-1}x+C     …(4)

Using the given condition \displaystyle y=1 when \displaystyle x=0,

\displaystyle 1=\frac12\log1+\frac34\log1-\frac12\tan^{-1}0+C

Since \displaystyle \log1=0 and \displaystyle \tan^{-1}0=0,

\displaystyle C=1

Substituting this value of C in (4),

\displaystyle y=\frac12\log(x+1)+\frac34\log(x^2+1)-\frac12\tan^{-1}x+1

Using the properties \displaystyle n\log a=\log(a^n) and \displaystyle \log a+\log b=\log(ab), and taking \frac14 common from first two terms, we get

\displaystyle y=\frac14\log\!\left((x+1)^2(x^2+1)^3\right)-\frac12\tan^{-1}x+1

Hence, the required particular solution is

\boxed{\displaystyle y=\frac14\log\!\left[(x+1)^2(x^2+1)^3\right]-\frac12\tan^{-1}x+1}

12. Find the particular solution of the following differential equation satisfying the given condition:

\displaystyle x(x^2-1)\frac{dy}{dx}=1,\; y=0\text{ when }x=2

Solution

Think First… After separating the variables, multiply the numerator and denominator by x. Then substitute t=x^2. This converts the denominator into the product of only two linear factors, making the partial fraction decomposition much easier.

Given, \displaystyle x(x^2-1)\frac{dy}{dx}=1

Separating the variables,

\displaystyle dy=\frac{dx}{x(x^2-1)}

Multiplying the numerator and denominator by x,

\displaystyle dy=\frac{x\,dx}{x^2(x^2-1)}

Integrating both sides, we get

\displaystyle \int dy=\int\frac{x\,dx}{x^2(x^2-1)}

Let \displaystyle t=x^2

Then, \displaystyle dt=2x\,dx\quad\Rightarrow\quad x\,dx=\frac{dt}{2}.

Therefore,

\displaystyle \int dy=\frac12\int\frac{dt}{t(t-1)}     …(1)

Using partial fractions on RHS of (1), we have

\displaystyle \frac1{t(t-1)}=\frac At+\frac B{t-1}

Multiplying both sides by \displaystyle t(t-1), we get

\displaystyle 1=A(t-1)+Bt

Expanding the right side,

\displaystyle 1=(A+B)t-A

Comparing the coefficients of like powers of t and constant terms, we get

\displaystyle A+B=0,\qquad -A=1 \Rightarrow A=-1

\displaystyle \therefore B=1

Thus,

\displaystyle \frac1{t(t-1)}=-\frac1t+\frac1{t-1}

Hence, equation (1) becomes

\displaystyle \int dy=\frac12\int\left(-\frac1t+\frac1{t-1}\right)dt

Using the standard integral

\displaystyle \int\frac{dx}{x}=\log|x|+C,

we get

\displaystyle y=\frac12\left[-\log|t|+\log|t-1|\right]+C

Substituting \displaystyle t=x^2, we get

\displaystyle y=\frac12\left[-\log(x^2)+\log|x^2-1|\right]+C

Using the property \displaystyle \log a-\log b=\log\left(\frac{a}{b}\right),

\displaystyle y=\frac12\log\left(\frac{x^2-1}{x^2}\right)+C     …(2)

Using the given condition \displaystyle y=0 when \displaystyle x=2,

\displaystyle 0=\frac12\log\left(\frac{2^2-1}{2^2}\right)+C

\displaystyle =\frac12\log\left(\frac34\right)+C

Therefore, \displaystyle C=-\frac12\log\left(\frac34\right)

Substituting this value of C in (2), we get

\displaystyle y=\frac12\log\left(\frac{x^2-1}{x^2}\right)-\frac12\log\left(\frac34\right)

Hence, the required particular solution is

\boxed{\displaystyle y=\frac12\log\left(\frac{x^2-1}{x^2}\right)-\frac12\log\left(\frac34\right)}

13. Find the particular solution of the following differential equation satisfying the given condition:

\displaystyle \cos\left(\frac{dy}{dx}\right)=a,\; (a\in\mathbb R);\; y=1\text{ when }x=0

Solution

Think First… The derivative is inside the cosine function. First apply the inverse cosine on both sides to obtain \displaystyle \frac{dy}{dx}. Since \displaystyle a is a constant, the right side becomes a constant and can be integrated directly.

Given \displaystyle \cos\left(\frac{dy}{dx}\right)=a

Taking inverse cosine on both sides, we get

\displaystyle \frac{dy}{dx}=\cos^{-1}a

Since \displaystyle a is a constant, \displaystyle \cos^{-1}a is also a constant.

Integrating both sides with respect to x, we get

\displaystyle \int dy=\int\cos^{-1}a\,dx

\displaystyle y=x\cos^{-1}a+C

Using the given condition \displaystyle y=1 when \displaystyle x=0,

\displaystyle 1=0\cdot\cos^{-1}a+C

\displaystyle \therefore C=1

Substituting this value of C, we get

\displaystyle y=x\cos^{-1}a+1

or \displaystyle y-1=x\cos^{-1}a

Dividing both sides by x,

\displaystyle \frac{y-1}{x}=\cos^{-1}a

Taking cosine on both sides, we get

\displaystyle \cos\left(\frac{y-1}{x}\right)=a

Hence, the required particular solution is

\boxed{\displaystyle \cos\left(\frac{y-1}{x}\right)=a}

14. Find the particular solution of the following differential equation satisfying the given condition:

\displaystyle \frac{dy}{dx}=y\tan x,\quad y=1\text{ when }x=0

Solution

Think First… Separate the variables by keeping all the y terms on the left side and all the x terms on the right side. Then use the standard integral of \tan x to obtain the general solution.

Given \displaystyle \frac{dy}{dx}=y\tan x

Separating the variables,

\displaystyle \frac{dy}{y}=\tan x\,dx

Integrating both sides,

\displaystyle \int\frac{dy}{y}=\int\tan x\,dx

Using the standard integrals

\displaystyle \int\frac{dx}{x}=\log|x|+C,\qquad \int\tan x\,dx=\log|\sec x|+C,

we get

\displaystyle \log|y|=\log|\sec x|+\log C

Using the property \displaystyle \log a+\log b=\log(ab),

\displaystyle \log|y|=\log|C\sec x|

\displaystyle \therefore y=C\sec x

Using the given condition \displaystyle y=1 when \displaystyle x=0,

\displaystyle 1=C\sec0=C \Rightarrow C=1

Substituting this value of C, we get

\displaystyle y=\sec x

Hence, the required particular solution is

\boxed{\displaystyle y=\sec x}

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is \displaystyle y'=e^x\sin x.

Solution

Think First… The differential equation already gives \displaystyle \frac{dy}{dx}. So, simply integrate both sides with respect to x. Since the curve passes through a given point, use that condition to determine the constant of integration.

Given,

\displaystyle \frac{dy}{dx}=e^x\sin x

Integrating both sides with respect to x, we get

\displaystyle \int dy=\int e^x\sin x\,dx

Using the integration by parts, we discussed in Exercise 7.6, we get

\displaystyle y=\frac{e^x}{2}(\sin x-\cos x)+C

Now, using the given condition that the curve passes through \displaystyle (0,0),

\displaystyle 0=\frac{e^0}{2}(\sin0-\cos0)+C

\displaystyle 0=-\frac12+C \Rightarrow C=\frac12

Substituting this value of C, we get

\displaystyle y=\frac{e^x}{2}(\sin x-\cos x)+\frac12

Rearranging the terms,

\displaystyle 2y-1=e^x(\sin x-\cos x)

Hence, the required equation of the curve is

\boxed{\displaystyle 2y-1=e^x(\sin x-\cos x)}

Maths Better Tip! Whenever the question asks for the equation of a curve passing through a given point, first obtain the general solution by integration and then use the given point to determine the constant of integration.

16. For the differential equation \displaystyle xy\frac{dy}{dx}=(x+2)(y+2), find the solution curve passing through the point \displaystyle (1,-1).

Solution

Think First… Separate the variables by collecting all the y terms on the left side and all the x terms on the right side. After integration, use the given point to determine the constant of integration.

Given,

\displaystyle xy\frac{dy}{dx}=(x+2)(y+2)

Separating the variables,

\displaystyle \frac{y}{y+2}\,dy=\frac{x+2}{x}\,dx

Writing each side in a simpler form,

\displaystyle \left(1-\frac2{y+2}\right)dy=\left(1+\frac2x\right)dx

Integrating both sides,

\displaystyle \int\left(1-\frac2{y+2}\right)dy=\int\left(1+\frac2x\right)dx

Using the standard integrals

\displaystyle \int dx=x+C,\qquad \int\frac{dx}{x}=\log|x|+C,

we get

\displaystyle y-2\log|y+2|=x+2\log|x|+C

Using the given condition that the curve passes through \displaystyle (1,-1),

\displaystyle -1-2\log|-1+2|=1+2\log1+C

Since \displaystyle \log1=0,

\displaystyle -1=1+C \Rightarrow C=-2

Substituting this value of C, we get

\displaystyle y-2\log|y+2|=x+2\log|x|-2

Rearranging the terms,

\displaystyle y-x+2=2\log|x|+2\log|y+2|

Using the property \displaystyle \log a^2=2\log a and \displaystyle \log a+\log b=\log(ab), we get

\displaystyle y-x+2=\log\!\left(x^2(y+2)^2\right)

Hence, the required solution curve is

\boxed{\displaystyle y-x+2=\log\!\left(x^2(y+2)^2\right)}

Maths Better Tip! Before integrating, rewrite rational expressions such as \displaystyle \frac{y}{y+2} and \displaystyle \frac{x+2}{x} into simpler forms by long division or factorising. This often reduces the integration to standard logarithmic integrals.

17. Find the equation of a curve passing through the point (0, –2) given that at any point \displaystyle (x,y) on the curve, the product of the slope of its tangent and the y-coordinate of the point is equal to the x-coordinate of the point.

Solution

Think First… First convert the given statement into a differential equation. Here, the product of the slope of the tangent and the y-coordinate equals the x-coordinate. Once the differential equation is formed, separate the variables, integrate, and use the given point to determine the constant of integration.

Let the slope of the tangent at any point \displaystyle (x,y) be \displaystyle \frac{dy}{dx}.

According to the given condition,

\displaystyle y\frac{dy}{dx}=x

Separating the variables,

\displaystyle y\,dy=x\,dx

Integrating both sides,

\displaystyle \int y\,dy=\int x\,dx

Using the standard integral

\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad n\neq-1,

we get

\displaystyle \frac{y^2}{2}=\frac{x^2}{2}+C

Multiplying both sides by 2,

\displaystyle y^2=x^2+2C

Using the given condition that the curve passes through \displaystyle (0,-2),

\displaystyle (-2)^2=0^2+2C

\displaystyle 4=2C

Substituting this value of 2C, we get

\displaystyle y^2=x^2+4

Rearranging the terms,

\displaystyle y^2-x^2=4

Hence, the required equation of the curve is

\boxed{\displaystyle y^2-x^2=4}

Maths Better Tip! In word problems, first translate the given statement into a differential equation. Once the equation is formed correctly, the remaining steps usually involve the standard variable separable method.

18. At any point \displaystyle (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point \displaystyle (-4,-3). Find the equation of the curve given that it passes through \displaystyle (-2,1).

Solution

Think First… First express the slope of the line joining the variable point \displaystyle (x,y) and the fixed point \displaystyle (-4,-3). Since the slope of the tangent is twice this slope, form the differential equation, separate the variables, and then use the given point to determine the constant of integration.

Let \displaystyle P(x,y) be any point on the curve.

The slope of the line joining \displaystyle P(x,y) and \displaystyle (-4,-3) is

\displaystyle \frac{y-(-3)}{x-(-4)}=\frac{y+3}{x+4}

According to the given condition, the slope of the tangent is twice this slope.

\displaystyle \therefore \frac{dy}{dx}=2\left(\frac{y+3}{x+4}\right)

Separating the variables,

\displaystyle \frac{dy}{y+3}=2\frac{dx}{x+4}

Integrating both sides,

\displaystyle \int\frac{dy}{y+3}=2\int\frac{dx}{x+4}

Using the standard integral

\displaystyle \int\frac{dx}{x}=\log|x|+C,

we get

\displaystyle \log|y+3|=2\log|x+4|+\log C

Using the properties \displaystyle 2\log a=\log a^2 and \displaystyle \log a+\log b=\log(ab), we get

\displaystyle \log|y+3|=\log\!\left(C(x+4)^2\right)

\displaystyle \therefore y+3=C(x+4)^2\qquad ...(1)

Using the given condition that the curve passes through \displaystyle (-2,1), we have

\displaystyle 1+3=C(-2+4)^2

\displaystyle 4=4C \Rightarrow C=1

Substituting this value of C in (1), we get

\displaystyle y+3=(x+4)^2

Hence, the required equation is

\boxed{\displaystyle (x+4)^2=y+3}

Maths Better Tip! In geometry-based differential equation problems, first express the required slope using the slope formula. Once the differential equation is formed correctly, the remaining steps usually involve the standard variable separable method.

19. The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Solution

Think First… The volume of a sphere depends on its radius. Differentiate the volume formula with respect to time to obtain a differential equation in \displaystyle r and \displaystyle t. Then separate the variables, integrate, and use the given conditions to determine the constant.

The volume of a sphere is given by

\displaystyle V=\frac43\pi r^3

Since the volume changes at a constant rate, let

\displaystyle \frac{dV}{dt}=k,\qquad\text{where }k\text{ is a constant.}

Differentiating \displaystyle V=\frac43\pi r^3 with respect to t, we get

\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

Therefore, we get

\displaystyle 4\pi r^2\frac{dr}{dt}=k

Separating the variables,

\displaystyle r^2\,dr=\frac{k}{4\pi}\,dt

Integrating both sides,

\displaystyle \int r^2\,dr=\frac{k}{4\pi}\int dt

Using the standard integrals

\displaystyle \int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\qquad \int dx=x+C,

we get

\displaystyle \frac{r^3}{3}=\frac{k}{4\pi}t+C

Multiplying both sides by 3,

\displaystyle r^3=\frac{3k}{4\pi}t+C

Using the initial condition that initially the radius is \displaystyle 3, that is, \displaystyle r=3 when \displaystyle t=0, we get

\displaystyle 27=C

Therefore,

\displaystyle r^3=\frac{3k}{4\pi}t+27\qquad ...(1)

Using the second condition that \displaystyle r=6 when \displaystyle t=3, in equation (1)

We get, \displaystyle 216=\frac{9k}{4\pi}+27

\displaystyle \frac{9k}{4\pi}=189

\displaystyle \frac{3k}{4\pi}=63

Substituting this value into equation (1), we get

\displaystyle r^3=63t+27

Taking cube root on both sides,

\displaystyle r=(63t+27)^{\frac13}

Hence, the required radius after t seconds is

\boxed{\displaystyle r=(63t+27)^{\frac13}}

Maths Better Tip! To determine all the arbitrary constants in the solution of a differential equation, you need the same number of independent conditions as the number of constants. Each condition helps eliminate one arbitrary constant.

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 doubles itself in 10 years. (Given: \displaystyle \log_e2=0.6931)

Solution

Think First… When the rate of increase is proportional to the principal, the differential equation is \displaystyle \frac{dP}{dt}=kP. First obtain the general solution, then use the given information to determine the growth rate.

Let \displaystyle P be the principal after t years.

Since the principal increases continuously at the rate of \displaystyle r\% per year,

\displaystyle \frac{dP}{dt}=\frac{r}{100}P

Separating the variables,

\displaystyle \frac{dP}{P}=\frac{r}{100}\,dt

Integrating both sides,

\displaystyle \int\frac{dP}{P}=\frac{r}{100}\int dt

Using the standard integrals

\displaystyle \int\frac{dx}{x}=\log|x|+C,\qquad \int dx=x+C,

we get

\displaystyle \log P=\frac{r}{100}t+C

Taking exponential on both sides,

\displaystyle P=Ae^{\frac{rt}{100}},\qquad\text{where }A=e^C\qquad ...(1)

Initially, the principal is Rs 100. Therefore, when \displaystyle t=0, equation (1) gives

\displaystyle 100=Ae^0 \Rightarrow A=100

Hence, \displaystyle P=100e^{\frac{rt}{100}}

21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs. 1000 is deposited with this bank. How much will it be worth after 10 years? (Given: \displaystyle e^{0.5}=1.648)

Solution

Think First… Since the principal increases continuously at a rate proportional to itself, form the differential equation \displaystyle \frac{dP}{dt}=kP. Then use the given initial amount to determine the constant of integration and evaluate the amount after 10 years.

Let \displaystyle P be the principal after t years.

Since the principal increases continuously at the rate of 5% per year,

\displaystyle \frac{dP}{dt}=\frac5{100}P=\frac1{20}P

Separating the variables,

\displaystyle \frac{dP}{P}=\frac1{20}\,dt

Integrating both sides,

\displaystyle \int\frac{dP}{P}=\frac1{20}\int dt

Using the standard integrals

\displaystyle \int\frac{dx}{x}=\log|x|+C,\qquad \int dx=x+C,

we get

\displaystyle \log P=\frac{t}{20}+C

Using \displaystyle \log_eP=u\Rightarrow P=e^u, we get

\displaystyle P=Ae^{\frac{t}{20}},\qquad\text{where }A=e^C\qquad ...(1)

Initially, the principal is Rs 1000. Therefore, when \displaystyle t=0, equation (1) gives

\displaystyle 1000=Ae^0 \Rightarrow A=1000

Hence, \displaystyle P=1000e^{\frac{t}{20}}

Now, After 10 years, \displaystyle t=10.

\displaystyle \therefore P=1000e^{\frac{10}{20}}=1000e^{0.5}

Using \displaystyle e^{0.5}=1.648,

\displaystyle P=1000\times1.648=1648

Hence, the amount after 10 years is

\boxed{\displaystyle \text{Rs. }1648}

Maths Better Tip! Questions involving continuous compound growth always lead to an exponential model. Once the growth equation is obtained, substitute the required time directly to calculate the final amount.

22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Solution

Think First… Since the rate of growth is proportional to the number of bacteria present, the differential equation is \displaystyle \frac{dN}{dt}=kN. First determine the constant of proportionality using the given 10% growth in 2 hours, and then find the time required for the population to double.

Let \displaystyle N be the number of bacteria after t hours.

Since the rate of growth is proportional to the number of bacteria present,

\displaystyle \frac{dN}{dt}=kN,\qquad\text{where }k\text{ is a constant.}

Separating the variables,

\displaystyle \frac{dN}{N}=k\,dt

Integrating both sides,

\displaystyle \int\frac{dN}{N}=k\int dt

Using the standard integrals

\displaystyle \int\frac{dx}{x}=\log|x|+C,\qquad \int dx=x+C,

we get

\displaystyle \log N=kt+C

Using \displaystyle \log_eN=u\Rightarrow N=e^u, we get

\displaystyle N=Ae^{kt},\qquad\text{where }A=e^C\qquad ...(1)

Initially, the bacteria count is \displaystyle 1,00,000. Therefore, when \displaystyle t=0, equation (1) gives

\displaystyle 100000=Ae^0 \Rightarrow A=100000

Hence, \displaystyle N=100000e^{kt}\qquad ...(2)

After 2 hours, the count increases by 10%. Therefore,

\displaystyle N=110000\qquad\text{when}\qquad t=2

Substituting these values in equation (2), we get

\displaystyle 110000=100000e^{2k}

or \displaystyle 1.1=e^{2k}

Taking natural logarithm on both sides,

\displaystyle \log(1.1)=2k

Now let the required time be \displaystyle T hours.

When the bacteria count becomes \displaystyle 200000, from equation (2), we get

\displaystyle 200000=100000e^{kT}

or \displaystyle 2=e^{kT}

Taking natural logarithm on both sides,

\displaystyle \log2=kT\qquad ...(3)

Using \displaystyle \log(1.1)=2k,

\displaystyle k=\frac{\log(1.1)}2

Substituting this value of k in equation (3), we get

\displaystyle T=\frac{2\log2}{\log(1.1)}=\frac{2\log2}{\log\left(\frac{11}{10}\right)}

Hence, the bacteria count will reach \displaystyle 2,00,000 after

\boxed{\displaystyle \frac{2\log2}{\log\left(\frac{11}{10}\right)}\text{ hours}}

Maths Better Tip! In exponential growth problems, use one known change to determine the growth constant first. Then use the same model to predict the population at any future time.

23. The general solution of the differential equation

\displaystyle \frac{dy}{dx}=e^{x+y} is

• (A) \displaystyle e^x+e^{-y}=C
• (B) \displaystyle e^x+e^y=C
• (C) \displaystyle e^{-x}+e^y=C
• (D) \displaystyle e^{-x}+e^{-y}=C

Solution

Think First… The given differential equation is of variable separable type. Separate the variables, integrate both sides, and then compare the resulting equation with the given options.

Given, \displaystyle \frac{dy}{dx}=e^{x+y}=e^x\cdot e^y

Separating the variables,

\displaystyle e^{-y}\,dy=e^x\,dx

Integrating both sides,

\displaystyle \int e^{-y}\,dy=\int e^x\,dx

Using the standard integrals

\displaystyle \int e^{ax}\,dx=\frac{e^{ax}}{a}+C,\qquad a\neq0,

we get

\displaystyle -e^{-y}=e^x+C

Adding \displaystyle e^{-y} to both sides,

\displaystyle 0=e^x+e^{-y}+C

Let \displaystyle C'=-C. Then,

\displaystyle e^x+e^{-y}=C'

Therefore, the general solution is

\boxed{\displaystyle e^x+e^{-y}=C'}

✅ Hence, the correct answer is (A)