NCERT Solutions: Class 12 Maths Chapter 9 - Differential Equations

The highest order derivative present is $\dfrac{d^4y}{dx^4}$, so the order is $4$. Since another derivative $y'''$ occurs inside the trigonometric function $\sin(y''')$, the equation is not a polynomial equation in derivatives. Hence its degree is not defined.

The highest order derivative is $y'$, so the order is $1$. The equation is polynomial in $y'$ and the highest power of $y'$ is $1$, so the degree is $1$.

The highest order derivative present is $\dfrac{d^2s}{dt^2}$, so the order is $2$. The equation is polynomial in derivatives, and the highest order derivative occurs to the first power. Hence the degree is $1$.

The highest order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$. Since $\dfrac{dy}{dx}$ occurs inside $\cos\left(\dfrac{dy}{dx}\right)$, the equation is not polynomial in derivatives. Therefore its degree is not defined.

The highest order derivative is $\dfrac{d^2y}{dx^2}$, so the order is $2$. This derivative occurs to the first power and the equation is polynomial in derivatives. Hence the degree is $1$.

The highest order derivative is $y'''$, so the order is $3$. The equation is polynomial in derivatives, and the highest order derivative $y'''$ occurs as $(y''')^2$. Therefore the degree is $2$.

The highest order derivative is $y'''$, so the order is $3$. It appears to the first power in a polynomial equation in derivatives, so the degree is $1$.

Only the first derivative $y'$ appears, so the order is $1$. It occurs to the first power in a polynomial equation in derivatives. Hence the degree is $1$.

The highest order derivative is $y''$, so the order is $2$. The equation is polynomial in derivatives, and $y''$ occurs to the first power. Hence the degree is $1$.

The highest order derivative is $y''$, so the order is $2$. The trigonometric term is a function of $y$, not of a derivative, so the equation remains polynomial in derivatives. Since $y''$ occurs to the first power, the degree is $1$.

1. i. $3$
2. ii. $2$
3. iii. $1$
4. iv. not defined

The derivative $\dfrac{dy}{dx}$ occurs inside $\sin\left(\dfrac{dy}{dx}\right)$. Therefore the equation is not polynomial in derivatives, so its degree is not defined.

1. i. $2$
2. ii. $1$
3. iii. $0$
4. iv. not defined

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

For $y=e^x+1$, $y'=e^x$ and $y''=e^x$. Hence $y''-y'=e^x-e^x=0$. Therefore the function is a solution.

Differentiating, $y'=2x+2$. Substitution gives $y'-2x-2=(2x+2)-2x-2=0$. Hence the function is a solution.

Differentiating, $y'=-\sin x$. Therefore $y'+\sin x=-\sin x+\sin x=0$. Hence the function is a solution.

For $y=(x^2+1)^{1/2}$, $y'=\dfrac{x}{\sqrt{x^2+1}}$. Also $\dfrac{xy}{x^2+1}=\dfrac{x\sqrt{x^2+1}}{x^2+1}=\dfrac{x}{\sqrt{x^2+1}}$. Both sides are equal.

Differentiating $y=Ax$, we get $y'=A$. Hence $xy'=xA=Ax=y$. Thus the differential equation is satisfied.

For $y=x\sin x$, $y'=\sin x+x\cos x$, so $xy'=x\sin x+x^2\cos x=y+x^2\cos x$. Also $x^2-y^2=x^2-x^2\sin^2x=x^2\cos^2x$. Under the stated condition, $\sqrt{x^2-y^2}=x\cos x$, so $y+x\sqrt{x^2-y^2}=y+x^2\cos x=xy'$.

Differentiating $xy=\log y+C$ with respect to $x$, we get $y+xy'=\dfrac{y'}{y}$. Thus $y'\left(\dfrac1y-x\right)=y$, so $y'=\dfrac{y}{\frac1y-x}=\dfrac{y^2}{1-xy}$. Hence the equation is satisfied.

Differentiating $y-\cos y=x$, we get $y'+\sin y\,y'=1$, so $y'=\dfrac1{1+\sin y}$. Since $x=y-\cos y$, $y\sin y+\cos y+x=y\sin y+\cos y+y-\cos y=y(1+\sin y)$. Therefore $(y\sin y+\cos y+x)y'=y(1+\sin y)\dfrac1{1+\sin y}=y$.

Differentiating $x+y=\tan^{-1}y$, we get $1+y'=\dfrac{y'}{1+y^2}$. Multiplying by $1+y^2$ gives $(1+y^2)+(1+y^2)y'=y'$, so $1+y^2+y^2y'=0$. Hence the differential equation is satisfied.

For $y=(a^2-x^2)^{1/2}$, $\dfrac{dy}{dx}=\dfrac{-x}{\sqrt{a^2-x^2}}=-\dfrac{x}{y}$. Hence $x+y\dfrac{dy}{dx}=x+y\left(-\dfrac{x}{y}\right)=0$. Thus the equation is satisfied.

1. i. $0$
2. ii. $2$
3. iii. $3$
4. iv. $4$

The general solution of a differential equation contains as many arbitrary constants as its order. For fourth order, the number is $4$.

1. i. $3$
2. ii. $2$
3. iii. $1$
4. iv. $0$

A particular solution is obtained after assigning definite values to the arbitrary constants. Hence it contains no arbitrary constant, regardless of the order.

Using $1-\cos x=2\sin^2\dfrac{x}{2}$ and $1+\cos x=2\cos^2\dfrac{x}{2}$, the equation becomes $\dfrac{dy}{dx}=\tan^2\dfrac{x}{2}$. Hence $y=\int\tan^2\dfrac{x}{2}\,dx=\int\left(\sec^2\dfrac{x}{2}-1\right)dx=2\tan\dfrac{x}{2}-x+C$.

Separate the variables: $\dfrac{dy}{\sqrt{4-y^2}}=dx$. Integrating both sides gives $\sin^{-1}\left(\dfrac{y}{2}\right)=x+C$.

The equation gives $\dfrac{dy}{dx}=1-y$. Separating, $\dfrac{dy}{1-y}=dx$. Integrating gives $-\log|1-y|=x+C_1$, or $\log|1-y|=-x+C$.

Rearrange as $\sec^2y\tan x\,dy=-\sec^2x\tan y\,dx$. Dividing by $\tan x\tan y$ gives $\dfrac{\sec^2y}{\tan y}\,dy=-\dfrac{\sec^2x}{\tan x}\,dx$. Hence $\log|\tan y|=-\log|\tan x|+C$, so $\tan x\tan y=C_1$.

The equation gives $dy=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\,dx$. Since $\dfrac{d}{dx}\log(e^x+e^{-x})=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}$, integration gives $y=\log(e^x+e^{-x})+C$.

Separate variables: $\dfrac{dy}{1+y^2}=(1+x^2)dx$. Integrating, $\tan^{-1}y=x+\dfrac{x^3}{3}+C$.

From $y\log y\,dx=x\,dy$, we get $\dfrac{dy}{y\log y}=\dfrac{dx}{x}$. Integrating, $\log|\log y|=\log|x|+C$, so $\log y=C_1x$.

Separate variables: $y^{-5}dy=-x^{-5}dx$. Integrating, $-\dfrac{1}{4y^4}=\dfrac{1}{4x^4}+C_1$. Multiplying by $-4$ and absorbing constants gives $\dfrac{1}{x^4}+\dfrac{1}{y^4}=C$.

Integrating, $y=\int\sin^{-1}x\,dx$. By parts, with $u=\sin^{-1}x$ and $dv=dx$, $\int\sin^{-1}x\,dx=x\sin^{-1}x-\int\dfrac{x}{\sqrt{1-x^2}}dx=x\sin^{-1}x+\sqrt{1-x^2}+C$.

Write $(1-e^x)\sec^2y\,dy=-e^x\tan y\,dx$. Then $\dfrac{\sec^2y}{\tan y}dy=\dfrac{e^x}{e^x-1}dx$. Integrating, $\log|\tan y|=\log|e^x-1|+C$, hence $\tan y=C_1(e^x-1)$.

Since $x^3+x^2+x+1=(x+1)(x^2+1)$, partial fractions give $\dfrac{2x^2+x}{x^3+x^2+x+1}=\dfrac{1}{2(x+1)}+\dfrac{3x-1}{2(x^2+1)}$. Hence $y=\dfrac12\log|x+1|+\dfrac34\log(x^2+1)-\dfrac12\tan^{-1}x+C$. Using $y=1$ at $x=0$ gives $C=1$.

Here $\dfrac{dy}{dx}=\dfrac{1}{x(x^2-1)}$. Since $\dfrac{1}{x(x^2-1)}=-\dfrac1x+\dfrac{1}{2(x-1)}+\dfrac{1}{2(x+1)}$, integration gives $y=-\log|x|+\dfrac12\log|x-1|+\dfrac12\log|x+1|+C$. Using $y=0$ at $x=2$ gives $C=\dfrac12\log\dfrac43$. Combining logs gives the stated solution.

Taking the principal value, $\dfrac{dy}{dx}=\cos^{-1}a$, a constant. Therefore $y=x\cos^{-1}a+C$. The condition $y=1$ when $x=0$ gives $C=1$.

Separating variables, $\dfrac{dy}{y}=\tan x\,dx$. Integrating gives $\log|y|=-\log|\cos x|+C=\log|\sec x|+C$. Thus $y=C_1\sec x$. Since $y=1$ at $x=0$, $C_1=1$ and $y=\sec x$.

Integrating, $y=\int e^x\sin x\,dx=\dfrac{e^x(\sin x-\cos x)}{2}+C$. The point $(0,0)$ gives $0=\dfrac{-1}{2}+C$, so $C=\dfrac12$. Hence $y=\dfrac{e^x(\sin x-\cos x)+1}{2}$.

Separate variables: $\dfrac{y}{y+2}dy=\dfrac{x+2}{x}dx$. Since $\dfrac{y}{y+2}=1-\dfrac{2}{y+2}$ and $\dfrac{x+2}{x}=1+\dfrac2x$, integration gives $y-2\log|y+2|=x+2\log|x|+C$. Substituting $(1,-1)$ gives $-1=1+C$, so $C=-2$.

The condition gives $y\dfrac{dy}{dx}=x$. Thus $y\,dy=x\,dx$. Integrating, $\dfrac{y^2}{2}=\dfrac{x^2}{2}+C$, or $y^2=x^2+C_1$. Since the curve passes through $(0,-2)$, $4=C_1$. Hence $y^2=x^2+4$.

The slope of the line joining $(x,y)$ to $(-4,-3)$ is $\dfrac{y+3}{x+4}$. Hence $\dfrac{dy}{dx}=\dfrac{2(y+3)}{x+4}$. Separating, $\dfrac{dy}{y+3}=2\dfrac{dx}{x+4}$. Integration gives $\log|y+3|=2\log|x+4|+C$, so $y+3=C_1(x+4)^2$. Using $(-2,1)$ gives $4=4C_1$, hence $C_1=1$.

The volume of a sphere is $V=\dfrac43\pi r^3$. Initially, $V_0=\dfrac43\pi(3)^3=36\pi$. After $3$ seconds, $V_3=\dfrac43\pi(6)^3=288\pi$. Since volume changes at a constant rate, $\dfrac{dV}{dt}=\dfrac{288\pi-36\pi}{3}=84\pi$. Hence $V=36\pi+84\pi t$. Therefore $\dfrac43\pi r^3=36\pi+84\pi t$, so $r^3=27+63t$ and $r=(27+63t)^{1/3}$.

For continuous growth, $\dfrac{dP}{dt}=\dfrac{r}{100}P$, so $P=Ce^{rt/100}$. With $P=100$ at $t=0$, $P=100e^{rt/100}$. Doubling in $10$ years gives $200=100e^{r/10}$, so $2=e^{r/10}$ and $r=10\log_e2=10(0.6931)=6.931$.

For continuous growth at $5\%$, $P=1000e^{0.05t}$. At $t=10$, $P=1000e^{0.5}=1000(1.648)=1648$. Hence the amount is Rs $1648$.

Let $N=N_0e^{kt}$. Since the count increases by $10\%$ in $2$ hours, $1.1N_0=N_0e^{2k}$, so $e^{2k}=1.1$ and $k=\dfrac12\log 1.1$. For the count to double, $2N_0=N_0e^{kt}$, so $t=\dfrac{\log2}{k}=\dfrac{2\log2}{\log1.1}\approx14.55$ hours.

1. i. $e^x+e^{-y}=C$
2. ii. $e^x+e^y=C$
3. iii. $e^{-x}+e^y=C$
4. iv. $e^{-x}+e^{-y}=C$

Since $\dfrac{dy}{dx}=e^xe^y$, separate variables: $e^{-y}dy=e^x dx$. Integrating, $-e^{-y}=e^x+C_1$, or $e^x+e^{-y}=C$. Hence option (i) is correct.

Here $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{x^2+xy}=\dfrac{1+(y/x)^2}{1+y/x}$, so it is homogeneous. Put $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. Then $x\dfrac{dv}{dx}=\dfrac{1-v}{1+v}$. Hence $\dfrac{1+v}{1-v}dv=\dfrac{dx}{x}$. Integrating, $-v-2\log|1-v|=\log|x|+C_1$, or $v+2\log|1-v|+\log|x|=C$. Substituting $v=\dfrac{y}{x}$ gives the result.

The right side is $1+\dfrac{y}{x}$, a function of $y/x$, so the equation is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=1+v$, so $x\dfrac{dv}{dx}=1$. Integrating gives $v=\log|x|+C$. Hence $y=x(\log|x|+C)$.

The equation gives $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}=\dfrac{1+y/x}{1-y/x}$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=\dfrac{1+v}{1-v}$, giving $\dfrac{1-v}{1+v^2}dv=\dfrac{dx}{x}$. Integrating, $\tan^{-1}v-\dfrac12\log(1+v^2)=\log|x|+C$. With $v=y/x$, this simplifies to $\tan^{-1}\left(\dfrac{y}{x}\right)-\dfrac12\log(x^2+y^2)=C$.

The equation gives $\dfrac{dy}{dx}=\dfrac{y^2-x^2}{2xy}=\dfrac{(y/x)^2-1}{2(y/x)}$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=\dfrac{v^2-1}{2v}$, hence $\dfrac{2v}{1+v^2}dv=-\dfrac{dx}{x}$. Integrating gives $\log(1+v^2)=-\log|x|+C$. Thus $x(1+y^2/x^2)=C_1$, or $x^2+y^2=Cx$.

The equation gives $\dfrac{dy}{dx}=1+\dfrac{y}{x}-2\left(\dfrac{y}{x}\right)^2$, a function of $y/x$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=1+v-2v^2$, so $x\dfrac{dv}{dx}=1-2v^2$. Separating, $\dfrac{dv}{1-2v^2}=\dfrac{dx}{x}$. Integrating gives $\dfrac{1}{2\sqrt2}\log\left|\dfrac{1+\sqrt2v}{1-\sqrt2v}\right|=\log|x|+C$. Substitute $v=y/x$.

Dividing by $dx$, $x\dfrac{dy}{dx}-y=\sqrt{x^2+y^2}$. Thus $\dfrac{dy}{dx}=\dfrac{y}{x}+\sqrt{1+(y/x)^2}$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=v+\sqrt{1+v^2}$, so $\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x}$. Integrating, $\log|v+\sqrt{1+v^2}|=\log|x|+C$. Substitute $v=y/x$.

The equation can be written with $v=y/x$ as $\dfrac{dy}{dx}=\dfrac{v(\cos v+v\sin v)}{v\sin v-\cos v}$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=\dfrac{v(\cos v+v\sin v)}{v\sin v-\cos v}$, so $x\dfrac{dv}{dx}=\dfrac{2v\cos v}{v\sin v-\cos v}$. Therefore $\left(\tan v-\dfrac1v\right)dv=2\dfrac{dx}{x}$. Integrating, $\log|\sec v|-\log|v|=2\log|x|+C$, so $\dfrac{\sec v}{v}=Cx^2$. With $v=y/x$, this gives $\sec(y/x)=Cxy$.

The equation gives $\dfrac{dy}{dx}=\dfrac{y}{x}-\sin\left(\dfrac{y}{x}\right)$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=v-\sin v$, giving $\dfrac{dv}{\sin v}=-\dfrac{dx}{x}$. Since $\int\csc v\,dv=\log|\csc v-\cot v|$, we get $\log|\csc v-\cot v|=-\log|x|+C$. Hence $\log|x(\csc v-\cot v)|=C$, and $v=y/x$.

The equation gives $\dfrac{dy}{dx}=-\dfrac{y}{x(\log(y/x)-2)}$, a function of $y/x$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=-\dfrac{v}{\log v-2}$, so $\dfrac{\log v-2}{v(1-\log v)}dv=\dfrac{dx}{x}$. Let $u=1-\log v$, so $du=-dv/v$. The left side becomes $\int\dfrac{1+u}{u}du=\log|u|+u$. Therefore $\log|1-\log v|+1-\log v=\log|x|+C$. Substitute $v=y/x$.

This equation is homogeneous in the form $dx/dy=h(x/y)$. Put $x=vy$, so $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$. Substituting gives $(1+e^v)\left(v+y\dfrac{dv}{dy}\right)+e^v(1-v)=0$, hence $y\dfrac{dv}{dy}(1+e^v)+v+e^v=0$. Thus $\dfrac{1+e^v}{v+e^v}dv=-\dfrac{dy}{y}$. Integrating, $\log|v+e^v|=-\log|y|+C$, or $y(v+e^v)=C$. Since $v=x/y$, the solution is $x+ye^{x/y}=C$.

The equation gives $\dfrac{dy}{dx}=\dfrac{y-x}{x+y}$, which is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=\dfrac{v-1}{1+v}$, so $\dfrac{1+v}{1+v^2}dv=-\dfrac{dx}{x}$. Integrating, $\tan^{-1}v+\dfrac12\log(1+v^2)=-\log|x|+C$. This simplifies to $\tan^{-1}(y/x)+\dfrac12\log(x^2+y^2)=C$. Using $(1,1)$ gives $C=\dfrac{\pi}{4}+\dfrac12\log2$.

The equation gives $\dfrac{dy}{dx}= -\dfrac{xy+y^2}{x^2}=-(v+v^2)$ with $v=y/x$, so it is homogeneous. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=-v-v^2$, hence $\dfrac{dv}{v(v+2)}=-\dfrac{dx}{x}$. Integrating, $\dfrac12\log\left|\dfrac{v}{v+2}\right|=-\log|x|+C$, so $\dfrac{v}{v+2}=\dfrac{C_1}{x^2}$. At $(1,1)$, $v=1$, so $C_1=\dfrac13$. Since $v=y/x$, this gives $\dfrac{y}{y+2x}=\dfrac1{3x^2}$, or $3x^2y=y+2x$.

The equation gives $x\dfrac{dy}{dx}=y-x\sin^2(y/x)$, so $\dfrac{dy}{dx}=v-\sin^2v$ where $v=y/x$. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=v-\sin^2v$, so $\csc^2v\,dv=-\dfrac{dx}{x}$. Integrating, $-\cot v=-\log|x|+C$, or $\cot v=\log|x|+C_1$. At $x=1$, $v=\pi/4$, so $C_1=1$. Hence $\cot(y/x)=\log|x|+1$.

The equation gives $\dfrac{dy}{dx}=\dfrac{y}{x}-\cosec(y/x)$. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=v-\cosec v$, so $\sin v\,dv=-\dfrac{dx}{x}$. Integrating, $-\cos v=-\log|x|+C$, or $\cos v=\log|x|+C_1$. Using $x=1,y=0$ gives $v=0$ and $C_1=1$. Hence $\cos(y/x)=\log|x|+1$.

The equation gives $\dfrac{dy}{dx}=\dfrac{2xy+y^2}{2x^2}=v+\dfrac{v^2}{2}$ where $v=y/x$. Put $y=vx$. Then $v+x\dfrac{dv}{dx}=v+\dfrac{v^2}{2}$, so $\dfrac{dv}{v^2}=\dfrac{dx}{2x}$. Integrating, $-\dfrac1v=\dfrac12\log|x|+C$. At $(1,2)$, $v=2$, giving $C=-\dfrac12$. Thus $\dfrac{x}{y}=\dfrac{1-\log|x|}{2}$, or $y=\dfrac{2x}{1-\log|x|}$.

1. i. $y=vx$
2. ii. $v=yx$
3. iii. $x=vy$
4. iv. $x=v$

Since the right hand side is a function of $x/y$, we put $x=vy$. Then $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$, which reduces the equation to a separable form.

1. i. $(4x+6y+5)dy-(3y+2x+4)dx=0$
2. ii. $(xy)dx-(x^3+y^3)dy=0$
3. iii. $(x^3+2y^2)dx+2xy\,dy=0$
4. iv. $y^2dx+(x^2-xy-y^2)dy=0$

A differential equation $Mdx+Ndy=0$ is homogeneous when $M$ and $N$ are homogeneous functions of the same degree. In option (iv), $M=y^2$ and $N=x^2-xy-y^2$ are both homogeneous of degree $2$. The other options contain constants or terms of different degrees. Hence option (iv) is correct.

This is linear with $P=2$ and $Q=\sin x$. The integrating factor is $e^{\int 2\,dx}=e^{2x}$. Hence $y e^{2x}=\int e^{2x}\sin x\,dx+C$. Since $\int e^{2x}\sin x\,dx=\dfrac{e^{2x}(2\sin x-\cos x)}{5}$, we get $y=\dfrac{2\sin x-\cos x}{5}+Ce^{-2x}$.

Here $P=3$, so the integrating factor is $e^{3x}$. Thus $y e^{3x}=\int e^{3x}e^{-2x}\,dx+C=\int e^x\,dx+C=e^x+C$. Therefore $y=e^{-2x}+Ce^{-3x}$.

The equation is linear with $P=1/x$. The integrating factor is $e^{\int (1/x)\,dx}=x$. Therefore $xy=\int x\cdot x^2\,dx+C=\dfrac{x^4}{4}+C$, and so $y=\dfrac{x^3}{4}+\dfrac{C}{x}$.

Here $P=\sec x$ and $Q=\tan x$. The integrating factor is $e^{\int \sec x\,dx}=\sec x+\tan x$. Hence $y(\sec x+\tan x)=\int \tan x(\sec x+\tan x)\,dx+C$. Since $\tan x\sec x+\tan^2x=\sec x\tan x+\sec^2x-1$, the integral is $\sec x+\tan x-x$. Thus $y(\sec x+\tan x)=\sec x+\tan x-x+C$.

Divide by $\cos^2x$ to get $\dfrac{dy}{dx}+y\sec^2x=\tan x\sec^2x$. The integrating factor is $e^{\int \sec^2x\,dx}=e^{\tan x}$. Therefore $y e^{\tan x}=\int \tan x\sec^2x e^{\tan x}\,dx+C$. Put $u=\tan x$; then the integral is $\int u e^u\,du=e^u(u-1)$. Hence $y=\tan x-1+Ce^{-\tan x}$.

Divide by $x$ to obtain $\dfrac{dy}{dx}+\dfrac{2}{x}y=x\log x$. The integrating factor is $e^{\int 2/x\,dx}=x^2$. Thus $x^2y=\int x^3\log x\,dx+C$. By parts, $\int x^3\log x\,dx=\dfrac{x^4}{4}\log x-\dfrac{x^4}{16}$. Hence $y=\dfrac{x^2}{4}\log x-\dfrac{x^2}{16}+\dfrac{C}{x^2}$.

Dividing by $x\log x$ gives $\dfrac{dy}{dx}+\dfrac{1}{x\log x}y=\dfrac{2}{x^2}$. The integrating factor is $e^{\int 1/(x\log x)\,dx}=\log x$. Thus $y\log x=\int \dfrac{2\log x}{x^2}\,dx+C$. Since $\int \dfrac{2\log x}{x^2}\,dx=-\dfrac{2(\log x+1)}{x}$, the result follows.

Writing the equation as $\dfrac{dy}{dx}+\dfrac{2x}{1+x^2}y=\dfrac{\cot x}{1+x^2}$, the integrating factor is $e^{\int 2x/(1+x^2)\,dx}=1+x^2$. Hence $(1+x^2)y=\int \cot x\,dx+C=\log|\sin x|+C$.

Divide by $x$ to get $\dfrac{dy}{dx}+\left(\dfrac1x+\cot x\right)y=1$. The integrating factor is $e^{\int(1/x+\cot x)\,dx}=x\sin x$. Therefore $xy\sin x=\int x\sin x\,dx+C=\sin x-x\cos x+C$.

Treat $x$ as a function of $y$. Since $\dfrac{dx}{dy}=x+y$, we have $\dfrac{dx}{dy}-x=y$. This linear equation has integrating factor $e^{-y}$. Thus $xe^{-y}=\int y e^{-y}\,dy+C=-(y+1)e^{-y}+C$. Multiplying by $e^y$ gives $x=Ce^y-y-1$.

Taking $x$ as a function of $y$, the equation becomes $y\dfrac{dx}{dy}+x-y^2=0$, or $\dfrac{dx}{dy}+\dfrac{x}{y}=y$. The integrating factor is $e^{\int (1/y)\,dy}=y$. Hence $xy=\int y^2\,dy+C=\dfrac{y^3}{3}+C$.

Write the equation in the form $\dfrac{dx}{dy}-\dfrac{x}{y}=3y$. The integrating factor is $e^{\int -1/y\,dy}=1/y$. Thus $\dfrac{x}{y}=\int 3\,dy+C=3y+C$, so $x=3y^2+Cy$.

The integrating factor is $e^{\int 2\tan x\,dx}=\sec^2x$. Thus $y\sec^2x=\int \sin x\sec^2x\,dx+C=\int \tan x\sec x\,dx+C=\sec x+C$. Hence $y=\cos x+C\cos^2x$. Using $y=0$ at $x=\pi/3$ gives $0=1/2+C/4$, so $C=-2$. Therefore $y=\cos x-2\cos^2x$.

Divide by $1+x^2$: $\dfrac{dy}{dx}+\dfrac{2x}{1+x^2}y=\dfrac{1}{(1+x^2)^2}$. The integrating factor is $1+x^2$. Therefore $(1+x^2)y=\int \dfrac{dx}{1+x^2}+C=\tan^{-1}x+C$. The condition $y=0$ at $x=1$ gives $C=-\pi/4$, so $(1+x^2)y=\tan^{-1}x-\dfrac{\pi}{4}$.

Here the integrating factor is $e^{\int -3\cot x\,dx}=\cosec^3x$. Hence $y\cosec^3x=\int \sin 2x\cosec^3x\,dx+C=\int 2\cot x\cosec x\,dx+C=-2\cosec x+C$. Thus $y=-2\sin^2x+C\sin^3x$. Using $y=2$ at $x=\pi/2$ gives $2=-2+C$, so $C=4$. Therefore $y=4\sin^3x-2\sin^2x$.

The condition gives $\dfrac{dy}{dx}=x+y$, or $\dfrac{dy}{dx}-y=x$. This linear equation has integrating factor $e^{-x}$. Hence $ye^{-x}=\int x e^{-x}\,dx+C=-(x+1)e^{-x}+C$. Therefore $y=Ce^x-x-1$. Since the curve passes through $(0,0)$, $0=C-1$, so $C=1$. Thus $y=e^x-x-1$.

The statement gives $x+y=\dfrac{dy}{dx}+5$, so $\dfrac{dy}{dx}-y=x-5$. This linear equation has integrating factor $e^{-x}$. Thus $ye^{-x}=\int (x-5)e^{-x}\,dx+C=(4-x)e^{-x}+C$. Hence $y=4-x+Ce^x$. Using $(0,2)$ gives $2=4+C$, so $C=-2$. Therefore $y=4-x-2e^x$.

1. i. $e^{-x}$
2. ii. $e^{-y}$
3. iii. $\dfrac{1}{x}$
4. iv. $x$

Dividing by $x$ gives $\dfrac{dy}{dx}-\dfrac{1}{x}y=2x$. Hence $P=-1/x$ and the integrating factor is $e^{\int -1/x\,dx}=e^{-\log x}=\dfrac1x$. Therefore the correct option is (iii).

1. i. $\dfrac{1}{y^2-1}$
2. ii. $\dfrac{1}{\sqrt{y^2-1}}$
3. iii. $\dfrac{1}{1-y^2}$
4. iv. $\dfrac{1}{\sqrt{1-y^2}}$

Dividing by $1-y^2$ gives $\dfrac{dx}{dy}+\dfrac{y}{1-y^2}x=\dfrac{ay}{1-y^2}$. Therefore $P_1=\dfrac{y}{1-y^2}$ and the integrating factor is $e^{\int y/(1-y^2)\,dy}=e^{-\frac12\log(1-y^2)}=\dfrac{1}{\sqrt{1-y^2}}$. Hence the correct option is (iv).

The order is the order of the highest derivative present. The degree is the power of the highest order derivative after the equation is polynomial in derivatives. In (i), the highest derivative is $d^2y/dx^2$ and it occurs to power $1$. In (ii), the highest derivative is $dy/dx$ and the highest power is $3$. In (iii), the highest derivative is $d^4y/dx^4$, so the order is $4$, but the equation contains $\sin(d^3y/dx^3)$ and is not polynomial in derivatives; hence the degree is not defined.

(i) From $xy=ae^x+be^{-x}+x^2$, differentiating gives $y+xy'=ae^x-be^{-x}+2x$ and differentiating again gives $2y'+xy''=ae^x+be^{-x}+2=xy-x^2+2$. Hence $xy''+2y'-xy+x^2-2=0$. (ii) Differentiating $y=e^x(a\cos x+b\sin x)$ twice gives $y''-2y'+2y=0$. (iii) If $y=x\sin 3x$, then $y'=\sin 3x+3x\cos 3x$ and $y''=6\cos 3x-9x\sin 3x$, so $y''+9y-6\cos 3x=0$. (iv) Differentiating $x^2=2y^2\log y$ gives $2x=2y(2\log y+1)y'$. Also $x^2+y^2=y^2(2\log y+1)$. Thus $(x^2+y^2)y'=xy$.

Differentiate $x^2-y^2=c(x^2+y^2)^2$ with respect to $x$: $2x-2yy'=4c(x^2+y^2)(x+yy')$. Since $c=\dfrac{x^2-y^2}{(x^2+y^2)^2}$, this becomes $(x-yy')(x^2+y^2)=2(x^2-y^2)(x+yy')$. Simplifying, $y'=\dfrac{x^3-3xy^2}{y^3-3x^2y}$. Hence $(x^3-3xy^2)dx=(y^3-3x^2y)dy$, proving the result.

The equation gives $\dfrac{dy}{dx}=-\sqrt{\dfrac{1-y^2}{1-x^2}}$. Separating variables, $\dfrac{dy}{\sqrt{1-y^2}}=-\dfrac{dx}{\sqrt{1-x^2}}$. Integrating, $\sin^{-1}y=-\sin^{-1}x+C$, or $\sin^{-1}y+\sin^{-1}x=C$.

Separating variables, $\dfrac{dy}{y^2+y+1}=-\dfrac{dx}{x^2+x+1}$. Therefore $\int \dfrac{dy}{y^2+y+1}+\int \dfrac{dx}{x^2+x+1}=C$. Completing squares and integrating gives $\tan^{-1}\left(\dfrac{2y+1}{\sqrt3}\right)+\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)=C_1$. Taking tangent of both sides gives $\dfrac{x+y+1}{1-x-y-2xy}=A$, where $A$ is a constant. Hence $(x+y+1)=A(1-x-y-2xy)$.

Divide by $\cos x\cos y$ to get $\tan x\,dx+\tan y\,dy=0$. Integrating, $-\log|\cos x|-\log|\cos y|=C$, or $\cos x\cos y=C_1$. Using the point $(0,\pi/4)$ gives $C_1=1\cdot \dfrac{1}{\sqrt2}=\dfrac{1}{\sqrt2}$. Hence $\cos x\cos y=\dfrac{1}{\sqrt2}$.

Write the equation as $\dfrac{dy}{1+y^2}=-\dfrac{e^x\,dx}{1+e^{2x}}$. Integrating gives $\tan^{-1}y=-\tan^{-1}(e^x)+C$, so $\tan^{-1}y+\tan^{-1}(e^x)=C$. At $x=0$, $y=1$, hence $C=\pi/4+\pi/4=\pi/2$.

Treat $x$ as a function of $y$ and put $x=vy$. Then $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$. Substitution gives $ye^v\left(v+y\dfrac{dv}{dy}\right)=vye^v+y^2$, so $y e^v\dfrac{dv}{dy}=1$. Hence $e^v\,dv=\dfrac{dy}{y}y=dy$. Integrating gives $e^v=y+C$. Since $v=x/y$, $e^{x/y}=y+C$.

Put $t=x-y$. Then $dt=dx-dy$ and the equation becomes $t(dx+dy)=dt$. Since $dy=dx-dt$, we have $dx+dy=2dx-dt$. Therefore $t(2dx-dt)=dt$, or $2t\,dx=(t+1)dt$. Thus $dx=\dfrac{t+1}{2t}dt$. Integrating, $x=\dfrac12(t+\log|t|)+C$. When $x=0,y=-1$, $t=1$, so $C=-1/2$. Hence $2x=t+\log|t|-1$, i.e. $2x=x-y+\log|x-y|-1$.

Invert the equation to write $\dfrac{dy}{dx}=\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}$, or $\dfrac{dy}{dx}+\dfrac{1}{\sqrt{x}}y=\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}$. The integrating factor is $e^{\int x^{-1/2}\,dx}=e^{2\sqrt{x}}$. Hence $y e^{2\sqrt{x}}=\int \dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}e^{2\sqrt{x}}\,dx+C=\int \dfrac{dx}{\sqrt{x}}+C=2\sqrt{x}+C$.

The integrating factor is $e^{\int \cot x\,dx}=\sin x$. Multiplying by it gives $\dfrac{d}{dx}(y\sin x)=4x$. Hence $y\sin x=2x^2+C$. Using $y=0$ at $x=\pi/2$ gives $0=\pi^2/2+C$, so $C=-\pi^2/2$. Therefore $y\sin x=2x^2-\dfrac{\pi^2}{2}$.

Put $u=e^y$. Then $du/dx=e^y dy/dx$. From $(x+1)dy/dx=2e^{-y}-1$, we get $(x+1)du/dx=2-u$, or $\dfrac{du}{2-u}=\dfrac{dx}{x+1}$. Integrating, $-\log|2-u|=\log|x+1|+C$, so $(x+1)(2-u)=C_1$. Since $u=e^y$ and $y=0$ when $x=0$, $C_1=1$. Thus $(x+1)(2-e^y)=1$.

1. i. $xy=C$
2. ii. $x=Cy^2$
3. iii. $y=Cx$
4. iv. $y=Cx^2$

The equation gives $y\,dx-x\,dy=0$, or $\dfrac{dy}{y}=\dfrac{dx}{x}$. Integrating, $\log|y|=\log|x|+C_1$, hence $y=Cx$. Therefore the correct option is (iii).

1. i. $y e^{\int P_1\,dy}=\int\left(Q_1e^{\int P_1\,dy}\right)dy+C$
2. ii. $y e^{\int P_1\,dx}=\int\left(Q_1e^{\int P_1\,dx}\right)dx+C$
3. iii. $x e^{\int P_1\,dy}=\int\left(Q_1e^{\int P_1\,dy}\right)dy+C$
4. iv. $x e^{\int P_1\,dx}=\int\left(Q_1e^{\int P_1\,dx}\right)dx+C$

For the linear equation $\dfrac{dx}{dy}+P_1x=Q_1$, the dependent variable is $x$ and the independent variable is $y$. Its integrating factor is $e^{\int P_1\,dy}$. Therefore $x e^{\int P_1\,dy}=\int\left(Q_1e^{\int P_1\,dy}\right)dy+C$, which is option (iii).

1. i. $xe^y+x^2=C$
2. ii. $xe^y+y^2=C$
3. iii. $ye^x+x^2=C$
4. iv. $ye^y+x^2=C$

Dividing by $dx$ and then by $e^x$ gives $\dfrac{dy}{dx}+y=-2xe^{-x}$. The integrating factor is $e^x$. Therefore $ye^x=\int -2x\,dx+C=-x^2+C$, or $ye^x+x^2=C$. Hence the correct option is (iii).