Differential Equations Class 12 Maths NCERT Solutions

Step-by-step solution

1. 1The highest order derivative present is $\frac{d^{4}y}{d{x}^4}$ — order 4.
2. 2The term $\sin \left(\frac{d^{3}y}{d{x}^3}\right)$ is transcendental in the derivative, so the equation is not a polynomial in derivatives — degree not defined.

Step-by-step solution

1. 1Highest order derivative present: $\frac{dy}{dx}$ — first order.
2. 2It appears to the power 1, so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $\frac{d^{2}s}{d{t}^2}$ — order 2.
2. 2It appears with power 1, so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $\frac{d^{2}y}{d{x}^2}$ — order 2.
2. 2The term $\cos \left(\frac{dy}{dx}\right)$ is transcendental in the derivative, so the equation is not a polynomial in derivatives — degree not defined.

Step-by-step solution

1. 1Highest order derivative present: $\frac{d^{2}y}{d{x}^2}$ — order 2.
2. 2It appears with power 1, so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $y^{\prime \prime \prime }$ = $\frac{d^{3}y}{d{x}^3}$ — order 3.
2. 2Its highest power in the equation is 2, so degree = 2.

Step-by-step solution

1. 1Highest order derivative present: $y^{\prime \prime \prime }$ — order 3.
2. 2It appears with power 1, so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $y^{\prime }$ — order 1.
2. 2It appears with power 1, so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $y^{\prime \prime }$ — order 2.
2. 2It appears with power 1, so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $y^{\prime \prime }$ — order 2.
2. 2The derivatives appear polynomially (sin y is not a function of a derivative), so degree = 1.

Step-by-step solution

1. 1Highest order derivative present: $\frac{d^{2}y}{d{x}^2}$ — order 2.
2. 2The term $\sin \left(\frac{dy}{dx}\right)$ makes the equation non-polynomial in the derivatives — degree not defined.

Step-by-step solution

1. 1Highest order derivative present: $\frac{d^{2}y}{d{x}^2}$ — order 2.
2. 2Square both sides to clear the fractional power: ${\left(\frac{d^{2}y}{d{x}^2}\right)}^4={\left[1+{\left(\frac{dy}{dx}\right)}^2\right]}^3$
3. 3The highest order derivative now appears with power 4, so degree = 4.

Step-by-step solution

1. 1y = eˣ + 1 gives y' = eˣ and y'' = eˣ.
2. 2$y^{\prime \prime }-y^{\prime }=e^x-e^x=0$ — an identity, so it is a solution.

Step-by-step solution

1. 1y' = 2x + 2.
2. 2$y^{\prime }-2x-2=(2x+2)-2x-2=0$ — identity.

Step-by-step solution

1. 1y' = −sin x.
2. 2$y^{\prime }+\sin x=-\sin x+\sin x=0$ — identity.

Step-by-step solution

1. 1y² = 1 + x², so 2yy' = 2x ⇒ y' = x/y.
2. 2$\frac{xy}{1+x^2}=\frac{xy}{y^2}=\frac{x}{y}=y^{\prime }$ — identity.

Step-by-step solution

1. 1y' = A.
2. 2$x{y}^{\prime }=xA=Ax=y$ — identity.

Step-by-step solution

1. 1y' = sin x + x cos x, so xy' = x sin x + x² cos x = y + x² cos x.
2. 2x² − y² = x² − x² sin²x = x² cos²x, so x√(x² − y²) = x² cos x (cos x > 0).
3. 3Hence xy' = y + x√(x² − y²) — identity. It is a solution.

Step-by-step solution

1. 1y² = a² − x² gives 2yy' = −2x ⇒ yy' = −x.
2. 2$x+y{y}^{\prime }=x-x=0$ — identity, valid for x ∈ (−a, a).

Step-by-step solution

1. 1y' = −sin x − cos x and y'' = −cos x + sin x.
2. 2$y^{\prime \prime }+y=(-\cos x+\sin x)+(\cos x-\sin x)=0$ — identity.

Step-by-step solution

1. 1Differentiate x + y = tan⁻¹y with respect to x: $1+y^{\prime }=\frac{y^{\prime }}{1+y^2}$
2. 2Multiply by 1 + y²: (1 + y²)(1 + y') = y' ⇒ 1 + y² + y²y' = 0 — identity.

Step-by-step solution

1. 1y' = −a sin x + b cos x, y'' = −a cos x − b sin x = −y.
2. 2$y^{\prime \prime }+y=0$ and the function carries two arbitrary constants — the general solution of this second order equation.

Step-by-step solution

1. 1The general solution of an equation of order n contains exactly n arbitrary constants.
2. 2Order 4 ⇒ 4 arbitrary constants.

Step-by-step solution

1. 1A particular solution is obtained from the general solution by fixing all the arbitrary constants using the initial conditions.
2. 2Hence it contains no arbitrary constants.

Step-by-step solution

1. 1The equation has two constants a and b, so differentiate twice.
2. 2First derivative: $\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0$
3. 3Second derivative: $\frac{1}{b}\frac{d^{2}y}{d{x}^2}=0\Rightarrow \frac{d^{2}y}{d{x}^2}=0$

Step-by-step solution

1. 1Differentiate once: 2yy' = −2ax ⇒ yy' = −ax.
2. 2Differentiate again: yy'' + (y')² = −a.
3. 3Substitute −a: from yy' = −ax we have a = −yy'/x, giving yy'' + (y')² = yy'/x.
4. 4Multiply by x: xyy'' + x(y')² − yy' = 0.

Step-by-step solution

1. 1Constant count 2 ⇒ differentiate twice.
2. 2$y^{\prime }=3a{e}^{3x}-2b{e}^{-2x},y^{\prime \prime }=9a{e}^{3x}+4b{e}^{-2x}$
3. 3Eliminate a, b: note 3y + y' = 6ae³ˣ and 3y' + y'' = 6be⁻²ˣ; multiplying the first relation of the system y'' − y' − 6y = 0 — verify: substitute both terms.
4. 4$y^{\prime \prime }-y^{\prime }-6y=(9a{e}^{3x}+4b{e}^{-2x})-(3a{e}^{3x}-2b{e}^{-2x})-6(a{e}^{3x}+b{e}^{-2x})=0$

Step-by-step solution

1. 1$y^{\prime }=e^{2x}(2a+b+2bx)=2y+b{e}^{2x},y^{\prime \prime }=4y+4b{e}^{2x}$
2. 2Eliminate be²ˣ = y' − 2y: y'' = 4y + 4(y' − 2y) = 4y' − 4y.

Step-by-step solution

1. 1$y^{\prime }=e^x(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x)=y+e^x(-a\sin x+b\cos x)$
2. 2$y^{\prime \prime }=y^{\prime }+e^x(-a\cos x-b\sin x)+e^x(-a\sin x+b\cos x)$
3. 3Eliminating gives y'' − 2y' + 2y = 0.

Step-by-step solution

1. 1Constants a, b, r (3) ⇒ differentiate thrice. First: $x-a+(y-b)y^{\prime }=0$
2. 2Second: $1+(y^{\prime }{)}^2+(y-b)y^{\prime \prime }=0$
3. 3$\Rightarrow y-b=\frac{1+(y^{\prime }{)}^2}{y^{\prime \prime }}$
4. 4Third/elimination leads to $(1+(y^{\prime }{)}^2{)}^3=r^2(y^{\prime \prime }{)}^2$ — substitute back: |1+(y')²| = r|y''|, square.

Step-by-step solution

1. 1Differentiate: $x{y}^{\prime }+y=a{e}^x-b{e}^{-x}$
2. 2Differentiate again: $x{y}^{\prime \prime }+2y^{\prime }=a{e}^x+b{e}^{-x}=xy$

Step-by-step solution

1. 1$y^{\prime }=-am\sin (mx)+bm\cos (mx),y^{\prime \prime }=-m^2(a\cos (mx)+b\sin (mx))=-m^{2}y$

Step-by-step solution

1. 1$y^{\prime }=-\frac{a}{x^2},y^{\prime \prime }=\frac{2a}{x^3}$
2. 2Eliminate a: from y' = −a/x², a/x³ = −y'/x, so y'' = −2y'/x ⇒ xy'' + 2y' = 0.

Step-by-step solution

1. 1Differentiate: 2x = 4ay' ⇒ 4a = 2x/y'.
2. 2Differentiate again: 2 = 4ay'' ⇒ substitute: 2 = (2x/y')y'' ⇒ xy'' − y' = 0.

Step-by-step solution

1. 1Differentiate: 2yy' = 4a.
2. 2Differentiate again: 2(y')² + 2yy'' = 0.

Step-by-step solution

1. 1For y = x: y' = 1, y'' = 0.
2. 2Check each: (A) 0 = x²? no. (B) 1 = x²? no. (C) 0 − x²(1) + x(x) = 0 ✓. (D) 0 = x³? no.

Step-by-step solution

1. 1$y^{\prime }=c_1{e}^x-c_2{e}^{-x},y^{\prime \prime }=c_1{e}^x+c_2{e}^{-x}=y$
2. 2Hence y'' − y = 0.

Step-by-step solution

1. 1Use identities: $\frac{1-\cos x}{1+\cos x}=\frac{2{\sin }^2(x/2)}{2{\cos }^2(x/2)}={\tan }^2(x/2)={\sec }^2(x/2)-1$
2. 2$y=\int ({\sec }^2(x/2)-1)dx=2\tan (x/2)-x+C$

Step-by-step solution

1. 1$\frac{dy}{\sqrt{4-y^2}}=dx\Rightarrow {\sin }^{-1}(y/2)=x+C$

Step-by-step solution

1. 1$\frac{dy}{y+1}=dx\Rightarrow \ln |y+1|=x+C$
2. 2y(0) = 5 ⇒ ln 6 = C, so $\ln \frac{y+1}{6}=x\Rightarrow y=6e^x-1$

Step-by-step solution

1. 1Separate: $\frac{{\sec }^{2}x}{\tan x}dx+\frac{{\sec }^{2}y}{\tan y}dy=0$
2. 2Integrate: ln|tan x| + ln|tan y| = C, so $\tan x\tan y=C$

Step-by-step solution

1. 1$dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\Rightarrow y=\ln (e^x+e^{-x})+C$

Step-by-step solution

1. 1$\frac{dy}{1+y^2}=(1+x^2)dx\Rightarrow {\tan }^{-1}y=x+\frac{x^3}{3}+C$

Step-by-step solution

1. 1$\frac{dy}{y\log y}=\frac{dx}{x}\Rightarrow \ln |\log y|=\ln |x|+C$
2. 2Hence $\log y=Cx\Rightarrow x=C\log y$

Step-by-step solution

1. 1$y^{-5}dy=-x^{-5}dx\Rightarrow \frac{y^{-4}}{-4}=\frac{x^{-4}}{4}+C$
2. 2Hence $\frac{1}{x^4}+\frac{1}{y^4}=C$

Step-by-step solution

1. 1$y=\int {\sin }^{-1}xdx=x{\sin }^{-1}x+\sqrt{1-x^2}+C$

Step-by-step solution

1. 1$y=\int {\tan }^{-1}xdx=x{\tan }^{-1}x-\frac{1}{2}\ln (1+x^2)+C$

Step-by-step solution

1. 1$dy=\frac{2x^2+x}{x^3+x^2+x+1}dx=\frac{2x^2+x}{(x+1)(x^2+1)}dx$
2. 2Partial fractions: $\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}$
3. 3Solving: A = 1/2, B = 3/2, C = −1/2.
4. 4$y=\frac{1}{2}\ln |x+1|+\frac{3}{4}\ln (x^2+1)-\frac{1}{2}{\tan }^{-1}x+C$

Step-by-step solution

1. 1$dy=\frac{dx}{x(x^2-1)}=\frac{dx}{x(x-1)(x+1)}$
2. 2Partial fractions: $\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$
3. 3Solving: A = −1, B = 1/2, C = 1/2.
4. 4$y=-\ln |x|+\frac{1}{2}\ln |x-1|+\frac{1}{2}\ln |x+1|+C=\frac{1}{2}\ln |x^2-1|-\ln |x|+C$

Step-by-step solution

1. 1$\frac{y}{\sqrt{1-y^2}}dy=-x{e}^{x}dx$
2. 2$-\sqrt{1-y^2}=-e^x(x-1)+C\Rightarrow e^x(x-1)-\sqrt{1-y^2}=C$

Step-by-step solution

1. 1$\frac{dy}{y}=2xdx\Rightarrow \ln |y|=x^2+C$
2. 2y(0) = 1 ⇒ C = 0, so $y=e^{x^2}$

Step-by-step solution

1. 1F(x,y) = (x² + y²)/(x² + xy) satisfies F(tx, ty) = F(x, y), so the equation is homogeneous; put y = vx, dy = v dx + x dv.
2. 2$(1+v)(vdx+xdv)=(1+v^2)dx\Rightarrow x(1+v)v^{\prime }=1-v$
3. 3$\frac{1+v}{1-v}dv=\frac{dx}{x}\Rightarrow -v-2\ln |1-v|=\ln |x|+C$
4. 4With y = vx: $\frac{y}{x}+\ln \left|\frac{(x-y{)}^2}{x}\right|=C$

Step-by-step solution

1. 1$\frac{x+y}{x}\text{ has degree 0, so it is homogeneous; put }y=vx:v+x{v}^{\prime }=1+v$
2. 2$x{v}^{\prime }=1\Rightarrow dv=\frac{dx}{x}\Rightarrow v=\ln |x|+C$
3. 3y = vx ⇒ $y=x\ln |x|+Cx$

Step-by-step solution

1. 1Put y = vx: $(1-v)(v+x{v}^{\prime })-(1+v)=0\Rightarrow x{v}^{\prime }=\frac{1+v^2}{1-v}$
2. 2$\frac{1-v}{1+v^2}dv=\frac{dx}{x}\Rightarrow {\tan }^{-1}v-\frac{1}{2}\ln (1+v^2)=\ln |x|+C$
3. 3Back-substitute v = y/x: $2{\tan }^{-1}(y/x)=\ln (x^2+y^2)+C$

Step-by-step solution

1. 1Put y = vx, dy = v dx + x dv: $(1-v^2)dx+2v(vdx+xdv)=0$
2. 2$(1+v^2)dx+2vxdv=0\Rightarrow \frac{dx}{x}+\frac{2v}{1+v^2}dv=0$
3. 3$\ln |x|+\ln (1+v^2)=C\Rightarrow x(1+v^2)=C$
4. 4y = vx ⇒ $x+\frac{y^2}{x}=C\Rightarrow x^2+y^2=Cx$

Step-by-step solution

1. 1Put y = vx: $v+x{v}^{\prime }=1+v-2v^2\Rightarrow x{v}^{\prime }=1-2v^2$
2. 2$\int \frac{dv}{1-2v^2}=\int \frac{dx}{x}\Rightarrow \frac{1}{2\sqrt{2}}\ln \left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=\ln |x|+C$
3. 3y = vx: $\frac{1}{2\sqrt{2}}\ln \left|\frac{x+\sqrt{2}y}{x-\sqrt{2}y}\right|=\ln |x|+C$

Step-by-step solution

1. 1Put y = vx, dy = v dx + x dv: $x(vdx+xdv)-vxdx=x\sqrt{1+v^2}dx$
2. 2$xdv=\sqrt{1+v^2}dx\Rightarrow \int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}$
3. 3$\ln |v+\sqrt{1+v^2}|=\ln |x|+C$
4. 4y = vx ⇒ $y+\sqrt{x^2+y^2}=C{x}^2$

Step-by-step solution

1. 1Put y = vx, dy = v dx + x dv: $x^{2}v(\cos v+v\sin v)dx=x^2(v\sin v-\cos v)(vdx+xdv)$
2. 2Cancel x² and collect dx terms: $2v\cos vdx=x(v\sin v-\cos v)dv$
3. 3$\frac{2}{x}dx=\left(\tan v-\frac{1}{v}\right)dv$
4. 4$2\ln |x|=-\ln |\cos v|-\ln |v|+C\Rightarrow x^{2}v\cos v=C$
5. 5v = y/x ⇒ $xy\cos (y/x)=C$

Step-by-step solution

1. 1$\frac{dy}{dx}=\frac{y}{x}-\sin (y/x)$
2. 2Put y = vx: $v+x{v}^{\prime }=v-\sin v\Rightarrow x{v}^{\prime }=-\sin v$
3. 3$\csc vdv=-\frac{dx}{x}\Rightarrow \ln |\tan (v/2)|=-\ln |x|+C$
4. 4y = vx ⇒ $x\tan \left(\frac{y}{2x}\right)=C$

Step-by-step solution

1. 1Write for x as a function of y: $\frac{dx}{dy}=\frac{x(2-\log (y/x))}{y}$
2. 2Put x = uy, dx/dy = u + y du/dy: $u+y\frac{du}{dy}=u(2-\log u)$
3. 3$y\frac{du}{dy}=u(1-\log u)\Rightarrow \frac{du}{u(1-\log u)}=\frac{dy}{y}$
4. 4Integrate (w = 1 − log u): $-\ln |1-\log u|=\ln |y|+C\Rightarrow y(1-\log u)=C$
5. 5u = x/y and 1 − log(x/y) = 1 + log(y/x): $1+\log \left(\frac{y}{x}\right)=\frac{C}{y}$

Step-by-step solution

1. 1Put x = uy, dx = u dy + y du: $(1+e^u)(udy+ydu)+e^u(1-u)dy=0$
2. 2$(u+e^u)dy+y(1+e^u)du=0\Rightarrow \frac{1+e^u}{u+e^u}du=-\frac{dy}{y}$
3. 3$\ln |u+e^u|=-\ln |y|+C\Rightarrow y(u+e^u)=C$
4. 4u = x/y ⇒ $x+y{e}^{x/y}=C$

Step-by-step solution

1. 1Put y = vx: $v+x{v}^{\prime }=v+v^2\Rightarrow \frac{dv}{v^2}=\frac{dx}{x}$
2. 2$-\frac{1}{v}=\ln |x|+C$
3. 3y(1) = 1 ⇒ v(1) = 1 ⇒ C = −1: $\frac{x}{y}=1-\ln |x|\Rightarrow y=\frac{x}{1-\ln |x|}$

Step-by-step solution

1. 1Put y = vx: $\cos v(v+x{v}^{\prime })=v\cos v-1\Rightarrow x\cos v{v}^{\prime }=-1$
2. 2$\cos vdv=-\frac{dx}{x}\Rightarrow \sin v=-\ln |x|+C$
3. 3y(1) = 0 ⇒ v(1) = 0 ⇒ C = 0: $\sin \left(\frac{y}{x}\right)=-\ln |x|$