Anand Classes brings you a comprehensive and free downloadable PDF of NCERT Solutions for Class 12 Maths – Chapter 9 “Differential Equations” (Miscellaneous Exercise Set-2), specially crafted to assist students in mastering advanced problems comprising the chapter’s miscellaneous exercise. These solutions follow the latest CBSE/NCERT syllabus meticulously and provide clear, step-by-step explanations for each question so that you can confidently review, revise and excel in board exams. Click the print button to download study material and notes.
NCERT Question.6 : Find the equation of the curve passing through the point $(0,\tfrac{\pi}{4})$ whose differential equation is
$$\sin x\cos ydx+\cos x\sin ydy=0.$$
Solution:
The given differential equation is
$$\sin x\cos ydx+\cos x\sin ydy=0.$$
Divide by $(\cos x\cos y)$:
$$\tan xdx+\tan ydy=0.$$
Integrate both sides:
$$\int \tan xdx+\int \tan ydy=\int0.$$
Using $\displaystyle\int \tan udu=\ln|\sec u|$ we get
$$\ln|\sec x|+\ln|\sec y|=\ln C.$$
$$\ln|\sec x\sec y|=\ln C.$$
Hence
$$\sec x\;sec y=C.$$
The curve passes through $(0,\tfrac{\pi}{4})$, so
$$\sec0\cdot\sec\frac{\pi}{4}=1\cdot\sqrt{2}=C.$$
Thus the required curve is
$$\sec x\;sec y=\sqrt{2}$$
equivalently
$$\cos y=\frac{1}{\sqrt{2}\cos x}=\frac{\sec x}{\sqrt{2}}\qquad\text{(for }\cos x\neq0\text{)}$$
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NCERT Question.7 : Find the particular solution of the differential equation
$$(1 + e^{2x})\;dy + (1 + y^{2})e^{x}\;dx = 0$$
given that $y=1$ when $x=0$.
Solution:
The given differential equation is
$$(1 + e^{2x})\;dy + (1 + y^{2})e^{x}\;dx = 0$$
Dividing every term by $(1 + y^2)(1 + e^{2x})$, we have
Rearrange as
$$\frac{dy}{1+y^{2}} + \frac{e^{x}\;dx}{1+e^{2x}} = 0.$$
Integrate both sides:
$$\int \frac{dy}{1+y^{2}} + \int \frac{e^{x}\;dx}{1+e^{2x}} = C.$$
Substitute $t=e^{x}$ so that $dt=e^{x}\;dx$ in the second integral:
$$\tan^{-1}y + \int \frac{dt}{1+t^{2}} = C.$$
Hence
$$\tan^{-1}y + \tan^{-1}(e^{x}) = C.$$
Apply the initial condition $x=0,\ y=1$:
$$\tan^{-1}1 + \tan^{-1}1 = C \implies \frac{\pi}{4} + \frac{\pi}{4} = C \implies C=\frac{\pi}{2}.$$
Thus the particular (implicit) solution is
$$\tan^{-1}y + \tan^{-1}(e^{x}) = \frac{\pi}{2}.$$
Solving explicitly,
$$\tan^{-1}y = \frac{\pi}{2} – \tan^{-1}(e^{x}) \implies y = \cot\big(\tan^{-1}(e^{x})\big) = \frac{1}{e^{x}} = e^{-x}.$$
Therefore the particular solution is
$$y = e^{-x}.$$
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NCERT Question.8 : Solve the differential equation
$$y\;e^{x/y}\;dx = \big(x\;e^{x/y} + y^2\big)\;dy\qquad(y\neq 0).$$
Solution:
Given differential equation
$$y\;e^{x/y}\;dx = \big(x\;e^{x/y} + y^2\big)\;dy\qquad(y\neq 0).$$
Rewrite the equation as
$$y\;e^{x/y}\;dx – x\;e^{x/y}\;dy = y^2\;dy.$$
Factor $(e^{x/y})$ on the left:
$$e^{x/y}\big(y\;dx – x\;dy\big) = y^2\;dy.$$
Divide both sides by $(y^2)$:
$$e^{x/y}\;\frac{y\;dx – x\;dy}{y^2} = dy.$$
Observe that
$$\frac{d}{dy}\big(e^{x/y}\big)=e^{x/y}\cdot\frac{d}{dy}\left(\frac{x}{y}\right)
= e^{x/y}\cdot\frac{y\;dx/dy – x}{y^2}
= e^{x/y}\;\frac{y\;dx – x\;dy}{dy\;y^2}.$$
Hence the left-hand side equals $(\dfrac{d}{dy}\big(e^{x/y}\big))$, so
$$\frac{d}{dy}\big(e^{x/y}\big)=1$$
Integrate with respect to (y):
$$e^{x/y}=y+C,$$
where (C) is an arbitrary constant.
Thus the general solution is
$$\boxed{e^{x/y}=y+C}.$$
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NCERT Question.9 : Find a particular solution of the differential equation
$$(x – y)(dx + dy) = dx – dy$$
given that $y=-1$ when $x=0$.
Solution:
Given differential equation
$$(x – y)(dx + dy) = dx – dy$$
Bring terms together:
$$(x-y-1)\;dx + (x-y+1)\;dy=0.$$
Thus
$$\frac{dy}{dx}=\frac{1-(x-y)}{1+(x-y)}.$$
Put $t=x-y$. Then
$$\frac{dt}{dx}=1-\frac{dy}{dx}.$$
Using the expression for $dy/dx$,
$$\frac{dt}{dx}=1-\frac{1-t}{1+t}=\frac{2t}{1+t}.$$
Separate variables:
$$\frac{1+t}{t}\;dt=2\;dx.$$
Integrate both sides:
$$\int\left(1+\frac{1}{t}\right)dt= \int 2\;dx$$
$$t+\ln|t|=2x+C.$$
Replace $t$ by $x-y$:
$$x-y+\ln|x-y|=2x+C.$$
Rearrange:
$$\ln|x-y|=x+y+C’.$$
Use the initial condition $x=0,\ y=-1$:
$$\ln|0-(-1)|=0+(-1)+C’ \implies \ln 1 = -1 + C’ \implies C’=1.$$
Hence the particular solution is
$$\boxed{\ln|x-y|=x+y+1}$$
(or equivalently $|x-y|=e^{x+y+1}$).
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NCERT Question.10 : Solve the differential equation
$$\left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1,\qquad x\neq0.$$
Solution:
Given differential equation
$$\left(\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right)\frac{dx}{dy}=1 $$
Rewrite the equation as
$$\frac{dx}{dy}=\frac{1}{\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}}.$$
Taking reciprocals gives
$$\frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}.$$
This is a linear first-order ODE of the form
$$\frac{dy}{dx}+\frac{1}{\sqrt{x}}y=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.$$
The integrating factor is
$$\mu(x)=\exp\left(\int\frac{1}{\sqrt{x}}dx\right)=\exp(2\sqrt{x})=e^{2\sqrt{x}}.$$
Multiply through by $\mu$:
$$\frac{d}{dx}\big(ye^{2\sqrt{x}}\big)=\frac{1}{\sqrt{x}}.$$
Integrate both sides:
$$ye^{2\sqrt{x}}=\int\frac{1}{\sqrt{x}}dx + C=2\sqrt{x}+C.$$
Thus the general solution is
$$y=e^{-2\sqrt{x}}\big(2\sqrt{x}+C\big),\qquad C\ \text{constant}.$$
Verification (optional): differentiating $y=e^{-2\sqrt{x}}(2\sqrt{x}+C)$ yields the original ODE after simplification.
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NCERT Question.11 : Find a particular solution of the differential equation
$$\left(\frac{dy}{dx}\right) + y \cot x = 4x \csc x, \quad (x \neq 0)$$
given that $y = 0$ when $x = \dfrac{\pi}{2}.$
Solution:
Given differential equation:
$$\left(\frac{dy}{dx}\right) + y \cot x = 4x \csc x$$
The given equation is a linear differential equation of the form
$$\left(\frac{dy}{dx}\right) + Py = Q$$
where
$$P = \cot x \quad \text{and} \quad Q = 4x \csc x.$$
Now, the Integrating Factor (I.F) is given by:
$$I.F = e^{\int P\;dx} = e^{\int \cot x\;dx} = e^{\log|\sin x|} = \sin x$$
Hence, the general solution of the given differential equation is:
$$y \cdot (I.F) = \int (Q \times I.F)\;dx + C$$
Substituting the values of $Q$ and $I.F$, we get
$$y \sin x = \int (4x \csc x \times \sin x)\;dx + C$$
Simplify the integrand:
$$y \sin x = 4 \int x\;dx + C$$
Integrating, we get
$$y \sin x = 4 \cdot \frac{x^2}{2} + C$$
$$y \sin x = 2x^2 + C$$
Now, using the given condition $x = \dfrac{\pi}{2}$ and $y = 0$, we find $C$:
$$0 = 2\left(\frac{\pi}{2}\right)^2 + C$$
$$0 = 2 \cdot \frac{\pi^2}{4} + C$$
$$0 = \frac{\pi^2}{2} + C$$
$$C = -\frac{\pi^2}{2}$$
Therefore,
$$y \sin x = 2x^2 – \frac{\pi^2}{2}.$$
Hence, the particular solution of the given differential equation
$$\left(\frac{dy}{dx}\right) + y \cot x = 4x \csc x$$
is
$$y \sin x = 2x^2 – \frac{\pi^2}{2}.$$
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NCERT Question.12 : Find a particular solution of the differential equation
$$(x+1)\frac{dy}{dx}=2e^{-y}-1,\qquad\text{given }y(0)=0.$$
Solution.
Given differential equation
$$(x+1)\frac{dy}{dx}=2e^{-y}-1$$
Rearrange to separate variables:
$$\frac{dy}{2e^{-y}-1}=\frac{dx}{x+1}.$$
Simplify the left integrand by letting $t=e^{y}$. Then $dt=e^{y}dy$, and note
$$2e^{-y}-1=\frac{2-e^{y}}{e^{y}}=\frac{2-t}{t}.$$
Hence
$$\frac{dy}{2e^{-y}-1}=\frac{tdy}{2-t}=\frac{dt}{2-t}.$$
Integrate both sides:
$$\int\frac{dt}{2-t}=\int\frac{dx}{x+1}.$$
The left integral gives $-\ln|2-t|$ and the right gives $\ln|x+1|+C$. Thus
$$-\ln|2-t|=\ln|x+1|+C.$$
Replace $t=e^{y}$:
$$-\ln|2-e^{y}|=\ln|x+1|+C.$$
Equivalently,
$$\ln|2-e^{y}|=-\ln|x+1|+C’$$
for some constant $C’$. Exponentiating,
$$|2-e^{y}|=\frac{C’}{x+1}.$$
Absorb sign into the constant and write
$$2-e^{y}=\frac{K}{x+1}.$$
Use the initial condition $x=0,\ y=0$: then $2-e^{0}=2-1=1$, so
$$1=\frac{K}{1}\implies K=1.$$
Therefore
$$2-e^{y}=\frac{1}{x+1}.$$
Solve for $e^{y}$:
$$e^{y}=2-\frac{1}{x+1}=\frac{2x+1}{x+1}.$$
Finally,
$$\boxed{y=\ln\left(\frac{2x+1}{x+1}\right),\qquad x\neq -1, \ \text{with } \frac{2x+1}{x+1}>0.,}$$
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NCERT Question.12 : The general solution of the differential equation
$$\frac{y\;dx – x\;dy}{y} = 0, \; is $$
(A) $xy = C \qquad (B) \;x = Cy^2 \qquad (C) \;y = Cx \qquad (D)\; y = Cx^2$
Solution:
Given differential equation:
$$\frac{y\;dx – x\;dy}{y} = 0$$
Simplify the equation:
$$\frac{ydx}{y} – \frac{xdy}{y} = 0$$
This gives:
$$dx = \frac{xdy}{y}$$
Rearrange:
$$\frac{dx}{x} = \frac{dy}{y}$$
Integrate both sides:
$$\int \frac{dx}{x} = \int \frac{dy}{y}$$
$$\ln |x| = \ln |y| + \ln C$$
Simplify:
$$\ln \left|\frac{x}{y}\right| = \ln C$$
Hence,
$$\frac{x}{y} = C \quad \text{or} \quad y = \frac{x}{C}$$
Letting $\dfrac{1}{C} = k$, we get
$$y = kx$$
Therefore, the general solution is
$$\boxed{y = Cx}$$
Correct Option: (C) $y = Cx$
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NCERT Question.13 : The general solution of a differential equation of the type
$${\frac{dx}{dy}} + P_1 x = Q_1$$
is:
(A) $y e^{\int P_1dy} = \int\big(Q_1 e^{\int P_1dy}\big)dy + C$
(B) $y e^{\int P_1dx} = \int\big(Q_1 e^{\int P_1dx}\big)dx + C$
(C) $x e^{\int P_1dy} = \int\big(Q_1 e^{\int P_1dy}\big)dy + C$
(D) $x e^{\int P_1dx} = \int\big(Q_1 e^{\int P_1dx}\big)dx + C$
Solution:
Write the equation in standard linear form for $x$ as a function of $y$:
$$\frac{dx}{dy}+P_1(y)x=Q_1(y).$$
The integrating factor is
$$\mu(y)=e^{\int P_1(y)dy}.$$
Multiplying through by $\mu(y)$ gives
$$\mu(y)\frac{dx}{dy}+\mu(y)P_1(y)x=\mu(y)Q_1(y)
\quad\Longrightarrow\quad\frac{d}{dy}\big(x\mu(y)\big)=\mu(y)Q_1(y).$$
Integrate with respect to $y$:
$$x\mu(y)=\int \mu(y)Q_1(y)dy + C.$$
Substitute $\mu(y)=e^{\int P_1dy}$ to obtain
$$\boxed{xe^{\int P_1,dy}=\int\big(Q_1e^{\int P_1dy}\big)dy + C}.$$
Correct option: (C).
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NCERT Question.14 : The general solution of the differential equation
$$e^xdy+(y e^x+2x)dx=0$$
is (choose):
(A) $x e^y + x^2 = C$
(B) $x e^y + y^2 = C$
(C) $y e^x + x^2 = C$
(D) $y e^y + x^2 = C$
Solution.
Given differential equation
$$e^xdy+(y e^x+2x)dx=0$$
Rewrite the equation as
$$e^x\frac{dy}{dx}+y e^x+2x=0.$$
Divide by $e^x$:
$$\frac{dy}{dx}+y=-2x e^{-x}.$$
This is linear with $P=1,;Q=-2x e^{-x}$.
Integrating factor $\mu(x)=e^{\int 1dx}=e^x$.
Multiply through by $\mu$:
$$e^x\frac{dy}{dx}+y e^x=\frac{d}{dx}\big(y e^x\big)=-2x.$$
Integrate:
$$y e^x=\int -2xdx + C = -x^2 + C.$$
Rearrange:
$$y e^x + x^2 = C.$$
Answer: (C) $y e^x + x^2 = C.$
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