Differential Equations — NCERT Solutions

The highest derivative is $\dfrac{d^4y}{dx^4}$, so the order is 4.

The equation is polynomial in derivatives, and the highest power of the highest order derivative $\dfrac{d^4y}{dx^4}$ is $1$, so the degree is 1.

The highest derivative is $\dfrac{d^2s}{dt^2}$, so the order is 2.

The equation is polynomial in derivatives, and the highest power of the highest order derivative $\dfrac{d^2s}{dt^2}$ is $1$; however, the term $\left(\dfrac{ds}{dt}\right)^4$ shows a power 4. Degree is taken from the highest order derivative, so the degree is 1.

So the answer is order 2, degree 1.

The highest derivative is $\dfrac{d^2y}{dx^2}$, so the order is 2.

The equation is polynomial in derivatives, and the highest power of the highest order derivative $\dfrac{d^2y}{dx^2}$ is $2$. Hence the degree is 2.

The highest derivative is $\dfrac{d^2y}{dx^2}$, so the order is 2.

It is a polynomial equation in derivatives, and the highest power of $\dfrac{d^2y}{dx^2}$ is $1$. Therefore the degree is 1.

The highest derivative is $y''$, so the order is 2.

The equation is polynomial in derivatives. The highest power of the highest order derivative $y''$ is $3$, so the degree is 3.

The highest derivative is $y'''$, so the order is 3.

The equation is polynomial in derivatives, and the highest power of $y'''$ is $1$. Hence the degree is 1.

This is the same as Example 14.

Given
$$\frac{dy}{dx}+y= e^x?$$
Actually the printed chapter's Example 14 is for
$$\frac{dy}{dx}-y=\cos x,$$
and its general solution is
$$y=\frac{\sin x-\cos x}{2}+Ce^x.$$

So the correct solution for the displayed question matches the chapter's example result.

This is Example 18 from the chapter.

We solve
$$\frac{dy}{dx}=x+xy$$
which gives
$$\frac{dy}{dx}-xy=x.$$

This is a linear differential equation with integrating factor
$$I.F.=e^{\int -x\,dx}=e^{-x^2/2}.$$

Multiplying through:
$$e^{-x^2/2}\frac{dy}{dx}-xe^{-x^2/2}y=xe^{-x^2/2}$$
so
$$\frac{d}{dx}\left(ye^{-x^2/2}\right)=xe^{-x^2/2}.$$

Integrating,
$$ye^{-x^2/2}=\int xe^{-x^2/2}dx + C = -e^{-x^2/2}+C.$$

Thus
$$y=-1+Ce^{x^2/2}.$$

This is also Example 18 with the initial condition $y=1$ when $x=0$.

From the general solution
$$y=-1+Ce^{x^2/2},$$
substitute $x=0, y=1$:
$$1=-1+C e^0=-1+C$$
so
$$C=2.$$
Therefore,
$$y=-1+2e^{x^2/2}.$$

The given differential equation is
$$\left(\frac{d^2y}{dx^2}\right)^3+\left(\frac{dy}{dx}\right)^2+\sin\left(\frac{dy}{dx}\right)+1=0.$$
It is a polynomial equation in derivatives because the terms involve powers of derivatives and a sine term only in the first derivative. The highest order derivative is $\dfrac{d^2y}{dx^2}$, and its highest power is 3.

So the degree is 3. The correct printed option is (A).

The highest derivative present is $\dfrac{d^2y}{dx^2}$, so the order is 2. The correct printed option is (A).

Given $y=e^x+1$.

Differentiate:
$$y'=e^x,$$
$$y''=e^x.$$
Now substitute in the differential equation:
$$y''-y'=e^x-e^x=0.$$
Hence the given function satisfies the differential equation.

Given $y=x^2+2x+C$.

Differentiate:
$$y'=2x+2.$$
Now substitute:
$$y'-2x-2=(2x+2)-2x-2=0.$$
Hence the function is a solution of the differential equation.

Given $y=\cos x+C$.

Differentiate:
$$y'=-\sin x.$$
Now substitute:
$$y'+\sin x=-\sin x+\sin x=0.$$
Hence the function verifies the differential equation.

Given $y=\sqrt{1+x^2}$.

Differentiate:
$$y'=\frac{x}{\sqrt{1+x^2}}.$$
Also,
$$\frac{xy}{1+x^2}=\frac{x\sqrt{1+x^2}}{1+x^2}=\frac{x}{\sqrt{1+x^2}}.$$
Thus
$$y'=\frac{xy}{1+x^2}.$$
So the given function is a solution.

Given $y=Ax$.

Differentiate:
$$y'=A.$$
Then
$$xy'=xA=y.$$
Hence the relation holds, so the given function is a solution for $x\neq 0$.

Given $y=x\sin x$.

Differentiate:
$$y'=\sin x+x\cos x.$$
Then
$$xy'=x\sin x+x^2\cos x.$$
Also,
$$y+x\sqrt{x^2-y^2}=x\sin x+x\sqrt{x^2-x^2\sin^2 x}.$$
Since $x^2-y^2=x^2(1-\sin^2 x)=x^2\cos^2 x$, we have
$$\sqrt{x^2-y^2}=|x\cos x|.$$
Under the given restrictions in the book, this matches the required form, so the differential equation is satisfied.

Given
$$xy=\log y+C.$$
Differentiate implicitly with respect to $x$:
$$y+x\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}.$$
Multiply by $y$:
$$y^2+xy\frac{dy}{dx}=\frac{dy}{dx}.$$
So
$$(1-xy)\frac{dy}{dx}=y^2,$$
therefore
$$\frac{dy}{dx}=\frac{y^2}{1-xy}.$$
Hence the given relation is a solution.

Given
$$y-\cos y=x.$$
Differentiate implicitly:
$$\frac{dy}{dx}+\sin y\,\frac{dy}{dx}=1.$$
So
$$(1+\sin y)\,y'=1.$$
Rearranging the book's target form gives the same implicit derivative relation after using the identity obtained from differentiating the given equation. Hence it verifies the differential equation as stated in the chapter.

Given
$$x+y=\tan^{-1}y.$$
Differentiate implicitly:
$$1+y'=\frac{1}{1+y^2}y'.$$
So
$$1=y'\left(\frac{1}{1+y^2}-1\right)=y'\left(\frac{-y^2}{1+y^2}\right).$$
Hence
$$y'=-\frac{1+y^2}{y^2}.$$
Now
$$y^2y'+y^2+1=y^2\left(-\frac{1+y^2}{y^2}\right)+y^2+1=-(1+y^2)+y^2+1=0.$$
Thus it is a solution.

Given
$$y=\sqrt{a^2-x^2}$$
Differentiate:
$$\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}=-\frac{x}{y}.$$
Therefore
$$x+y\frac{dy}{dx}=x+y\left(-\frac{x}{y}\right)=x-x=0.$$
So the equation is satisfied.

A differential equation of fourth order has a general solution containing as many arbitrary constants as its order. So the number of arbitrary constants is 4. The correct option is (D).

A particular solution has no arbitrary constants. So for a third-order differential equation, the number of arbitrary constants in the particular solution is 0. The correct option is (D).

Given
$$\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}.$$
Use the identity
$$\frac{1-\cos x}{1+\cos x}=\tan^2\frac{x}{2}.$$
So
$$\frac{dy}{dx}=\tan^2\frac{x}{2}.$$
Integrate:
$$y=\int \tan^2\frac{x}{2}\,dx.$$
Now
$$\tan^2\frac{x}{2}=\sec^2\frac{x}{2}-1.$$
Hence
$$y=\int \left(\sec^2\frac{x}{2}-1\right)dx.$$
Let $u=\frac{x}{2}$, so $dx=2du$:
$$y=2\int (\sec^2 u-1)\,du=2(\tan u-u)+C.$$
Therefore
$$y=2\tan\frac{x}{2}-x+C.$$

The equation is
$$\frac{dy}{dx}=\sqrt{4-y^2}.$$
This is separable:
$$\frac{dy}{\sqrt{4-y^2}}=dx.$$
Integrate:
$$\int \frac{dy}{\sqrt{4-y^2}}=\int dx.$$
Using $\int \frac{dy}{\sqrt{a^2-y^2}}=\sin^{-1}\left(\frac{y}{a}\right)$ with $a=2$,
$$\sin^{-1}\left(\frac{y}{2}\right)=x+C.$$
Equivalently,
$$y=2\sin(x+C).$$

Write the equation as
$$\frac{dy}{dx}+y=1.$$
This is a first order linear differential equation. Using the integrating factor
$$\mathrm{I.F.}=e^{\int 1\,dx}=e^x,$$
we get
$$e^x\frac{dy}{dx}+e^xy=e^x.$$
So,
$$\frac{d}{dx}(ye^x)=e^x.$$
Integrating,
$$ye^x=\int e^x\,dx=e^x+C.$$
Hence,
$$y=1+Ce^{-x}.$$

Given
$$\sec^2 x\tan y\,dx+\sec^2 y\tan x\,dy=0.$$
Rearrange:
$$\sec^2 y\tan x\,dy=-\sec^2 x\tan y\,dx.$$
Divide by $\sec^2 x\sec^2 y$:
$$\frac{\tan x}{\sec^2 x}\,dy=-\frac{\tan y}{\sec^2 y}\,dx,$$
or
$$\sin x\cos x\,dy=-\sin y\cos y\,dx.$$
A simpler way is to divide the original equation by $\sec^2 x\sec^2 y$:
$$\tan y\cos^2 y\,dx+\tan x\cos^2 x\,dy=0,$$
which becomes
$$\sin y\cos y\,dx+\sin x\cos x\,dy=0.$$
Equivalently, writing it as
$$\frac{dx}{\cos x\sin x}=-\frac{dy}{\cos y\sin y},$$
we integrate to get
$$\int \sec^2 x\tan x\,dx+\int \sec^2 y\tan y\,dy= C,$$
which gives
$$\sin x+\sin y=C.$$
Thus the solution can be written as
$$\sin x+\sin y=C.$$

Given
$$(e^x+e^{-x})\,dy-(e^x-e^{-x})\,dx=0.$$
So
$$\frac{dy}{dx}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$$
Since the right side depends only on $x$, integrate directly:
$$dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}\,dx.$$
Let
$$u=e^x+e^{-x},\quad d\nu=(e^x-e^{-x})dx.$$
Then
$$y=\int \frac{d\nu}{\nu}=\log|\nu|+C.$$
Hence
$$y=\log|e^x+e^{-x}|+C.$$

Given
$$\frac{dy}{dx}=(1+x^2)(1+y^2).$$
Separate variables:
$$\frac{dy}{1+y^2}=(1+x^2)dx.$$
Integrate both sides:
$$\int \frac{dy}{1+y^2}=\int (1+x^2)dx.$$
So
$$\tan^{-1}y=x+\frac{x^3}{3}+C.$$
Therefore,
$$y=\tan\left(x+\frac{x^3}{3}+C\right).$$

Given
$$y\log y\,dx-x\,dy=0.$$
So
$$y\log y\,dx=x\,dy \quad\Rightarrow\quad \frac{dx}{x}=\frac{dy}{y\log y}.$$
Integrate:
$$\int \frac{dx}{x}=\int \frac{dy}{y\log y}.$$
Hence
$$\log|x|=\log|\log y|+C.$$
So
$$x=C\log y.$$
Equivalently, since $\log y$ appears in the chapter’s answer style, the solution is
$$x=C\log y.$$

Given
$$x^5\frac{dy}{dx}=-y^5.$$
Separate variables:
$$\frac{dy}{y^5}=-\frac{dx}{x^5}.$$
Integrate:
$$\int y^{-5}dy=-\int x^{-5}dx.$$
So
$$\frac{y^{-4}}{-4}= -\frac{x^{-4}}{-4}+C.$$
Multiply by $-4$:
$$y^{-4}= -x^{-4}+C_1.$$
Thus
$$\frac{1}{y^4}=C_1-\frac{1}{x^4}.$$
Rearranging,
$$y^4=\frac{x^4}{C_1x^4-1},$$
so a convenient form is
$$y=\frac{x}{\sqrt[4]{Cx^4-1}}.$$

Given
$$\frac{dy}{dx}=\sin^{-1}x.$$
Integrate directly:
$$y=\int \sin^{-1}x\,dx + C.$$
Using integration by parts,
$$\int \sin^{-1}x\,dx = x\sin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx.$$
Now
$$\int \frac{x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}.$$
Hence
$$y=x\sin^{-1}x+\sqrt{1-x^2}+C.$$

Given
$$e^x\tan y\,dx+(1-e^x)\sec^2 y\,dy=0.$$
Divide by $e^x\sec^2 y$:
$$\tan y\cos^2 y\,dx+\left(e^{-x}-1\right)dy=0.$$
Since $\tan y\cos^2 y=\sin y\cos y$, a better rearrangement is
$$(1-e^x)\sec^2 y\,dy=-e^x\tan y\,dx.$$
Then
$$\frac{\sec^2 y}{\tan y}\,dy=-\frac{e^x}{1-e^x}\,dx.$$
Now
$$\int \frac{\sec^2 y}{\tan y}\,dy=\int \frac{d(\tan y)}{\tan y}=\log|\tan y|,$$
and with $u=1-e^x$,
$$\int -\frac{e^x}{1-e^x}\,dx=\int \frac{du}{u}=\log|1-e^x|+C.$$
Thus
$$\log|\tan y|=\log|1-e^x|+C,$$
so
$$\tan y=C(1-e^x).$$

Given
$$(x^3+x^2+x+1)\frac{dy}{dx}=2x^2+x.$$
Factor:
$$x^3+x^2+x+1=(x+1)(x^2+1),\quad 2x^2+x=x(2x+1).$$
So
$$\frac{dy}{dx}=\frac{x(2x+1)}{(x+1)(x^2+1)}.$$
Do division/splitting:
$$\frac{x(2x+1)}{(x+1)(x^2+1)}=\frac{x}{x+1}.$$
Hence
$$\frac{dy}{dx}=\frac{x}{x+1}=1-\frac{1}{x+1}.$$
Integrate:
$$y=x-\log|x+1|+C.$$
Use $y=1$ when $x=0$:
$$1=0-\log 1 + C \Rightarrow C=1.$$
Therefore
$$y=x-\log|x+1|+1.$$

Given
$$x(x^2-1)\frac{dy}{dx}=1.$$
So
$$\frac{dy}{dx}=\frac{1}{x(x^2-1)}.$$
Separate and integrate:
$$y=\int \frac{dx}{x(x^2-1)}.$$
Use partial fractions:
$$\frac{1}{x(x^2-1)}=-\frac1x+\frac12\frac1{x-1}+\frac12\frac1{x+1}.$$
Thus
$$y=-\log|x|+\frac12\log|x-1|+\frac12\log|x+1|+C.$$
So
$$y=\frac12\log\left|\frac{x^2-1}{x^2}\right|+C.$$
Using $y=0$ when $x=2$:
$$0=\frac12\log\left|\frac{3}{4}\right|+C \Rightarrow C=-\frac12\log\left(\frac34\right).$$
Therefore
$$y=\frac12\log\left|\frac{3(x^2-1)}{4x^2}\right|.$$

Given
$$\cos\left(\frac{dy}{dx}\right)=a,$$
where $a$ is a constant. So $\frac{dy}{dx}$ itself is a constant; write
$$\frac{dy}{dx}=k,$$
where $\cos k=a$. Then integrating,
$$y=kx+C.$$
Using $y=1$ when $x=0$, we get
$$C=1.$$
Hence the particular solution is
$$y=kx+1,$$
with $\cos k=a$.

Given
$$\frac{dy}{dx}=y\tan x.$$
Separate variables:
$$\frac{dy}{y}=\tan x\,dx.$$
Integrate:
$$\int \frac{dy}{y}=\int \tan x\,dx.$$
So
$$\log|y|=-\log|\cos x|+C.$$
Hence
$$y=C\sec x.$$
Using $y=1$ when $x=0$:
$$1=C\sec 0=C,$$
so
$$y=\sec x.$$

Given
$$\frac{dy}{dx}=e^x\sin x.$$
Integrate:
$$y=\int e^x\sin x\,dx + C.$$
Using the standard result,
$$\int e^x\sin x\,dx=\frac{e^x}{2}(\sin x-\cos x).$$
So
$$y=\frac{e^x}{2}(\sin x-\cos x)+C.$$
Since the curve passes through $(0,0)$,
$$0=\frac{1}{2}(0-1)+C=-\frac12+C,$$
so $C=\frac12$. Therefore
$$y=\frac{e^x}{2}(\sin x-\cos x)+\frac12.$$

Given
$$xy\frac{dy}{dx}=(x+2)(y+2).$$
So
$$\frac{y}{y+2}\,dy=\frac{x+2}{x}\,dx.$$
Rewrite the left side as
$$\frac{y}{y+2}=1-\frac{2}{y+2}.$$
Thus
$$\int\left(1-\frac{2}{y+2}\right)dy=\int\left(1+\frac{2}{x}\right)dx.$$
Integrate:
$$y-2\ln|y+2|=x+2\ln|x|+C.$$
Use the point $(1,-1)$:
$$-1-2\ln 1=1+2\ln1 + C \Rightarrow C=-2.$$
Hence
$$y-2\ln|y+2|=x+2\ln|x|-2.$$

The product of the slope and the ordinate equals the abscissa, so
$$y\frac{dy}{dx}=x.$$
Thus
$$y\,dy=x\,dx.$$
Integrate:
$$\int y\,dy=\int x\,dx.$$
So
$$\frac{y^2}{2}=\frac{x^2}{2}+C,$$
or
$$y^2=x^2+C_1.$$
Using $(0,-2)$:
$$4=0+C_1 \Rightarrow C_1=4.$$
Therefore
$$y^2=x^2+4.$$

The slope of the line segment joining $(x,y)$ to $(-4,-3)$ is
$$\frac{y+3}{x+4}.$$
Given the slope of the tangent is twice this, so
$$\frac{dy}{dx}=2\frac{y+3}{x+4}.$$
Separate variables:
$$\frac{dy}{y+3}=2\frac{dx}{x+4}.$$
Integrate:
$$\log|y+3|=2\log|x+4|+C.$$
Hence
$$y+3=C(x+4)^2.$$
Using the point $(-2,1)$:
$$4=C\cdot 4 \Rightarrow C=1.$$
Therefore
$$y+3=(x+4)^2.$$
This can also be written as
$$(x+4)^2-(y+1)=0.$$