NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3 - Differential Equations

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

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

$\\ (\int \frac{dy}{\sqrt{a^2-y^2}} = sin^{-1}\frac{y}{a})\\$

The required general solution:

$\\ \implies sin^{-1}\frac{y}{2} = x + C$

Question 3: Find the general solution: $\frac{dy}{dx} + y = 1 (y\neq 1)$

Answer:

Given, in the question

$\frac{dy}{dx} + y = 1$

$\\ \implies \frac{dy}{dx} = 1- y \\$

$\implies \int\frac{dy}{1-y} = \int dx$

$(\int\frac{dx}{x} = lnx)$

$\\ \implies -log(1-y) = x + C\ \ (We\ can\ write\ C= log k) \\$

$\implies log k(1-y) = -x \\$

$\implies 1- y = \frac{1}{k}e^{-x} \\$

The required general equation

$\implies y = 1 -\frac{1}{k}e^{-x}$

Question 4: Find the general solution: $\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

Answer:

Given,

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

$\\ \implies \frac{sec^2 y}{tan y}dy = -\frac{sec^2 x}{tan x}dx \\$

$\implies \int \frac{sec^2 y}{tan y}dy = - \int \frac{sec^2 x}{tan x}dx$

Now, let tany = t and tanx = u

$sec^2 y dy = dt\ and\ sec^2 x dx = du$

$\\ \implies \int \frac{dt}{t} = -\int \frac{du}{u} \\$

$\implies log t = -log u +logk \\ \implies t = \frac{1}{ku} \\$

$\implies tany = \frac{1}{ktanx}$

Question 5: Find the general solution:

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

Answer:

Given, in the question

$(e^x + e^{-x})dy - (e^x - e^{-x})dx = 0$

$\\ \implies dy = \frac{(e^x - e^{-x})}{(e^x + e^{-x})}dx$

Let,

$\\ (e^x + e^{-x}) = t \\$

$\implies (e^x - e^{-x})dx = dt$

$\\ \implies \int dy = \int \frac{dt}{t} \\$

$\implies y = log t + C \\ $

$\implies y = log(e^x + e^{-x}) + C$

This is the general solution

Question 6: Find the general solution: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

Answer:

Given, in the question

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

$\\ \implies \int \frac{dy}{(1+y^2)} = \int (1+x^2)dx$

$(\int \frac{dx}{(1+x^2)} =tan^{-1}x +c)$

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

Question 7: Find the general solution: $y\log y dx - x dy = 0$

Answer:

Given,

$y\log y dx - x dy = 0$

$\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx$

let logy = t

=> 1/ydy = dt

$\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx \\$

$\implies \log t = \log x + \log k \\$

$\implies t = kx \\$

$\implies \log y = kx$

This is the general solution

Question 8: Find the general solution: $x^5\frac{dy}{dx} = - y^5$

Answer:

Given, in the question

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

$\\ \implies \int \frac{dy}{y^5} = - \int \frac{dx}{x^5} \\$

$\implies \frac{y^{-4}}{-4} = -\frac{x^{-4}}{-4} + C \\$

$\implies \frac{1}{y^4} + \frac{1}{x^4} = C$

This is the required general equation.

Question 9: Find the general solution: $\frac{dy}{dx} = \sin^{-1}x$

Answer:

Given, in the question

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

$\implies \int dy = \int \sin^{-1}xdx$

Now,

$\int (u.v)dx = u\int vdx - \int(\frac{du}{dx}.\int vdx)dx$

Here, u = $\sin^{-1}x$ and v = 1

$\implies y = \sin^{-1}x .x - \int(\frac{1}{\sqrt{1-x^2}}.x)dx$

$\\ Let\ 1- x^2 = t \\$

$\implies -2xdx = dt \implies xdx = -dt/2$

$\\ \implies y = x\sin^{-1}x+ \int(\frac{dt}{2\sqrt{t}}) \\ $

$\implies y = x\sin^{-1}x + \frac{1}{2}.2\sqrt{t} + C \\$

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

Question 10: Find the general solution $e^x\tan y dx + (1-e^x)\sec^2 y dy = 0$

Answer:

Given,

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

$\\ \implies e^x\tan y dx = - (1-e^x)\sec^2 y dy \\ $

$\implies \int \frac{\sec^2 y }{\tan y}dy = -\int \frac{e^x }{(1-e^x)}dx$

$\\ let\ tany = t \ and \ 1-e^x = u \\$

$\implies \sec^2 ydy = dt\ and \ -e^xdx = du$

$\\ \therefore \int \frac{dt }{t} = \int \frac{du }{u} \\$

$\implies \log t = \log u + \log k \\$

$\implies t = ku \\$

$\implies \tan y= k (1-e^x)$

Question 11: Find a particular solution satisfying the given condition:

$(x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x; \ y = 1\ \textup{when}\ x = 0$

Answer:

Given, in the question

$
\begin{aligned}
& \left(x^3+x^2+x+1\right) \frac{d y}{d x}=2 x^2+x \\
& \Longrightarrow \int d y=\int \frac{2 x^2+x}{\left(x^3+x^2+x+1\right)} d x \\
& \left(x^3+x^2+x+1\right)=(x+1)\left(x^2+1\right)
\end{aligned}
$

Now,

$
\begin{aligned}
& \Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^2+1} \\
& \Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{A x^2+A(B x+C)(x+1)}{(x+1)\left(x^2+1\right)} \\
& \Rightarrow 2 x^2+x=A x^2+A+B x+C x+C \\
& \Rightarrow 2 x^2+x=(A+B) x^2+(B+C) x+A+C
\end{aligned}
$

Now comparing the coefficients

$
A+B=2 ; B+C=1 ; A+C=0
$

Solving these:

$
\mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{3}{2}, \mathrm{C}=-\frac{1}{2}
$

Putting the values of $A, B, C$ :

$
\Rightarrow \frac{2 x^2+x}{(x+1)\left(x^2+1\right)}=\frac{1}{2} \frac{1}{(x+1)}+\frac{1}{2} \frac{3 x-1}{x^2+1}
$

Therefore,

$
\begin{aligned}
& \Rightarrow \int d y=\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{3 x-1}{x^2+1} d x \\
& \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{2} \int \frac{\mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \int \frac{\mathrm{dx}}{\mathrm{x}^2+1} \\
& \Rightarrow \mathrm{y}=\frac{1}{2} \log (\mathrm{x}+1)+\frac{3}{4} \int \frac{2 \mathrm{x}}{\mathrm{x}^2+1} \mathrm{dx}-\frac{1}{2} \tan ^{-1} \mathrm{x}
\end{aligned}
$

Question 12: Find a particular solution satisfying the given condition:

$x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2$

Answer:

Given, in the question $x\left(x^2-1\right) \frac{d y}{d x}=1$

$
\begin{aligned}
& \Longrightarrow \int d y=\int \frac{d x}{x\left(x^2-1\right)} \\
& \Longrightarrow \int d y=\int \frac{d x}{x(x-1)(x+1)}
\end{aligned}
$

Let,

$
\begin{aligned}
& \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{c}{x-1} \\
& \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A(x-1)(x+1)+B(x)(x-1)+C(x)(x+1)}{x(x+1)(x-1)} \\
& \Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{(A+B+C) x^2+(B-C) x-A}{x(x+1)(x-1)}
\end{aligned}
$

Now comparing the values of $A, B, C$

$
A+B+C=0 ; B-C=0 ; A=-1
$

Solving these:

$
\mathrm{B}=\frac{1}{2} \text { and } \mathrm{C}=\frac{1}{2}
$

Now putting the values of $A, B, C$,

$
\begin{aligned}
& \Rightarrow \frac{1}{x(x+1)(x-1)}=-\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{x-1}\right) \\
& \Rightarrow \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int\left(\frac{1}{x+1}\right) d x+\frac{1}{2} \int\left(\frac{1}{x-1}\right) d x \\
& \Rightarrow y=-\log x+\frac{1}{2} \log (x+1)+\frac{1}{2} \log (x-1)+\log c \\
& \left.\Rightarrow \mathrm{y}=\frac{1}{2} \log \left[\frac{\mathrm{c}^2(\mathrm{x}-1)(\mathrm{x}+1)}{\mathrm{x}^2}\right\}----\mathrm{iii}\right)
\end{aligned}
$

Given, $\mathrm{y}=0$ when $\mathrm{x}=2$

$
\begin{aligned}
& 0=\frac{1}{2} \log \left[\frac{\mathrm{c}^2(2-1)(2+1)}{4}\right\} \\
& \Rightarrow \log \frac{3 \mathrm{c}^2}{4}=0
\end{aligned}
$

$
\Rightarrow 3 c^2=4
$

Therefore,

$
\begin{aligned}
& \Longrightarrow y=\frac{1}{2} \log \left[\frac{4(x-1)(x+1)}{3 x^2}\right] \\
& \Longrightarrow y=\frac{1}{2} \log \left[\frac{4\left(x^2-1\right)}{3 x^2}\right]
\end{aligned}
$

Question 13: Find a particular solution satisfying the given condition:

$\cos\left(\frac{dy}{dx} \right ) = a\ (a\in R);\ y = 1\ \textup{when}\ x = 0$

Answer:

Given,

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

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

$\implies \int dy = \int\cos^{-1}a\ dx \\$

$\implies y = x\cos^{-1}a + c$

Now, y =1 when x =0

1 = 0 + c

Therefore, c = 1

Putting the value of c:

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

Question 14: Find a particular solution satisfying the given condition:

$\frac{dy}{dx} = y\tan x; \ y =1\ \textup{when}\ x = 0$

Answer:

Given,

$\frac{dy}{dx} = y\tan x$

$\\ \implies \int \frac{dy}{y} = \int \tan x\ dx \\$

$\implies \log y = \log \sec x + \log k \\$

$\implies y = k\sec x$

Now, y=1 when x =0

1 = ksec0

$\implies$ k = 1

Putting the vlue of k:

y = sec x

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

Answer:

We first find the general solution of the given differential equation

Given,

$y' = e^x\sin x$

$\\ \implies \int dy = \int e^x\sin xdx$

$\\ Let I = \int e^x\sin xdx \\$

$\implies I = \sin x.e^x - \int(\cos x. e^x)dx \\$

$\implies I = e^x\sin x - [e^x\cos x - \int(-\sin x.e^x)dx] \\$

$\implies 2I = e^x(\sin x - \cos x) \\$

$\implies I = \frac{1}{2}e^x(\sin x - \cos x)$

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + c$

Now, Since the curve passes through (0,0)

y = 0 when x =0

$\\ \therefore 0 = \frac{1}{2}e^0(\sin 0 - \cos 0) + c \\$

$\implies c = \frac{1}{2}$

Putting the value of c, we get:

$\\ \therefore y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2} \\$

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

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

Answer:

We first find the general solution of the given differential equation

Given,

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

$\\ \implies \int \frac{y}{y+2}dy = \int \frac{x+2}{x}dx \\ $

$\implies \int \frac{(y+2)-2}{y+2}dy = \int (1 + \frac{2}{x})dx \\ $

$\implies \int (1 - \frac{2}{y+2})dy = \int (1 + \frac{2}{x})dx \\$

$\implies y - 2\log (y+2) = x + 2\log x + C$

Now, Since the curve passes through (1,-1)

y = -1 when x = 1

$\\ \therefore -1 - 2\log (-1+2) = 1 + 2\log 1 + C \\$

$\implies -1 -0 = 1 + 0 +C \\ \implies C = -2$

Putting the value of C:

$\\ y - 2\log (y+2) = x + 2\log x + -2 \\$

$\implies y -x + 2 = 2\log x(y+2)$

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

Answer:

According to the question,

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

$\\ \implies \int ydy =\int xdx \\$

$\implies \frac{y^2}{2} = \frac{x^2}{2} + c$

Now, Since the curve passes through (0,-2).

x =0 and y = -2

$\\ \implies \frac{(-2)^2}{2} = \frac{0^2}{2} + c \\ \implies c = 2$

Putting the value of c, we get

$\\ \frac{y^2}{2} = \frac{x^2}{2} + 2 \\ \implies y^2 = x^2 + 4$

Question 18: At any point (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 (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Answer:

Slope m of line joining (x,y) and (-4,-3) is $\frac{y+3}{x+4}$

According to the question,

$\\ \frac{dy}{dx} = 2(\frac{y+3}{x+4}) \\$

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

$\implies \log (y+3) = 2\log (x+4) + \log k \\$

$\implies (y+3) = k(x+4)^2$

Now, Since the curve passes through (-2,1)

x = -2 , y =1

$\\ \implies (1+3) = k(-2+4)^2 \\ \implies k =1$

Putting the value of k, we get

$\\ \implies y+3 = (x+4)^2$

Question 19: The volume of 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 balloon after t seconds.

Answer:

Volume of a sphere, $V = \frac{4}{3}\pi r ^3$

Given, Rate of change is constant.

$\\ \therefore \frac{dV}{dt} = c \\$

$\implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c \\ $

$\implies \int d(\frac{4}{3}\pi r ^3) = c\int dt \\$

$\implies \frac{4}{3}\pi r ^3 = ct + k$

Now, at t=0, r=3 and at t=3 , r =6

Putting these value:

$\frac{4}{3}\pi (3) ^3 = c(0) + k \\ \implies k = 36\pi$

Also,

$\frac{4}{3}\pi (6) ^3 = c(3) + 36\pi \\ $

$\implies 3c = 252\pi \\ \implies c = 84\pi$

Putting the value of c and k:

$\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\$

$\implies r ^3 = (21 t + 9)(3) = 62t + 27 \\$

$\implies r = \sqrt[3]{62t + 27}$

Question 20: In a bank, principal increases continuously at the rate of r % per year. Find the value of r if Rs 100 double itself in 10 years (log _e 2 = 0.6931).

Answer:

Let p be the principal amount and t be the time.

According to question,

$\frac{dp}{dt} = (\frac{r}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{r}{100})dt \\ $

$\implies \log p = \frac{r}{100}t + C$

$\\ \implies p = e^{\frac{rt}{100} + C}$

Now, at t =0 , p = 100

and at t =10, p = 200

Putting these values,

$\\ \implies 100 = e^{\frac{r(0)}{100} + C} = e^C$

Also,

, $\\ \implies 200 = e^{\frac{r(10)}{100} + C} = e^{\frac{r}{10}}.e^C = e^{\frac{r}{10}}.100 \\$

$\implies e^{\frac{r}{10}} = 2 \\$

$\implies \frac{r}{10} = \ln 2 = 0.6931 \\ $

$\implies r = 6.93$

So value of r = 6.93%

Question 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 worth after 10 years (e ^0.5 = 1.648).

Answer:

Let p be the principal amount and t be the time.

According to question,

$\frac{dp}{dt} = (\frac{5}{100})p$

$\\ \implies \int\frac{dp}{p} = \int (\frac{1}{20})dt \\$

$\implies \log p = \frac{1}{20}t + C$

$\\ \implies p = e^{\frac{t}{20} + C}$

Now, at t =0 , p = 1000

Putting these values,

$\\ \implies 1000 = e^{\frac{(0)}{20} + C} = e^C$

Also, At t=10 \,

$\\ \implies p = e^{\frac{(10)}{20} + C} = e^{\frac{1}{2}}.e^C = e^{\frac{1}{2}}.1000 \\$

$\implies p =(1.648)(1000) = 1648$

After 10 years, the total amount would be Rs.1648

Question 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?

Answer:

Let n be the number of bacteria at any time t.

According to question,

$\frac{dn}{dt} = kn\ \ (k\ is\ a\ constant)$

$\\ \implies \int \frac{dn}{n} = \int kdt \\$

$\implies \log n = kt + C$

Now, at t=0, n = 100000

$\\ \implies \log (100000) = k(0) + C \\$

$\implies C = 5$

Again, at t=2, n= 110000

$\\ \implies \log (110000) = k(2) + 5 \\$

$\implies \log 11 + 4 = 2k + 5 \\$

$\implies 2k = \log 11 -1 =\log \frac{11}{10} \\$

$\implies k = \frac{1}{2}\log \frac{11}{10}$

Using these values, for n= 200000

$\\ \implies \log (200000) = kt + C \\ $

$\implies \log 2 +5 = kt + 5 \\ \implies (\frac{1}{2}\log \frac{11}{10})t = \log 2 $

$\\ \implies t = \frac{2\log 2}{ \log \frac{11}{10}}$

Question 23: The general solution of the differential equation $\frac{dy}{dx} = e^{x+y}$ is

(A) $e^x + e^{-y} = C$

(B) $e^{x }+ e^{y} = C$

(C) $e^{-x }+ e^{y} = C$

(D) $e^{-x }+ e^{-y} = C$

Answer:

Given,

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

$\\ \implies\frac{dy}{dx} = e^x.e^y \\ \implies\int \frac{dy}{e^y} = \int e^x.dx \\$

$\implies -e^{-y} = e^x + C \\ \implies e^x + e^{-y} = K\ \ \ \ (Option A)$