Anand Classes provides a comprehensive, free-downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations, Miscellaneous Exercise (Set-1). This resource includes detailed, step-by-step answers to a wide range of questions covering concepts like order and degree of differential equations, general and particular solutions, formation of differential equations from given families of curves, separable and homogeneous first-order equations, and application-based problems. These solutions are aligned with the latest CBSE-NCERT syllabus and are ideal for thorough board exam preparation and revision. Click the print button to download study material and notes.
NCERT Question.1 : For each of the differential equations given below, indicate its order and degree (if defined)
(i) $\dfrac{d^2y}{dx^2} + 5x\left(\dfrac{dy}{dx}\right)^2 – 6y = \log x$
(ii) $\left(\dfrac{dy}{dx}\right)^3 – 4 \left(\dfrac{dy}{dx}\right)^2 + 7y = \sin x$
(iii) $\dfrac{d^4y}{dx^4} – \sin\left(\dfrac{d^3y}{dx^3}\right) = 0$
Solution:
Order and Degree of Differential Equations
(i) $\dfrac{d^2y}{dx^2} + 5x\left(\dfrac{dy}{dx}\right)^2 – 6y = \log x$
Rewriting the equation:
$$
\frac{d^2y}{dx^2} + 5x\left(\frac{dy}{dx}\right)^2 – 6y – \log x = 0
$$
- The highest order derivative present is $\dfrac{d^2y}{dx^2}$.
Order: 2 - The highest power of $\dfrac{d^2y}{dx^2}$ is 1.
Degree: 1
(ii) $\left(\dfrac{dy}{dx}\right)^3 – 4 \left(\dfrac{dy}{dx}\right)^2 + 7y = \sin x$
Rewriting the equation:
$$
\left(\frac{dy}{dx}\right)^3 – 4 \left(\frac{dy}{dx}\right)^2 + 7y – \sin x = 0
$$
- The highest order derivative present is $\dfrac{dy}{dx}$.
Order: 1 - The highest power of $\dfrac{dy}{dx}$ is 3.
Degree: 3
(iii) $\dfrac{d^4y}{dx^4} – \sin\left(\dfrac{d^3y}{dx^3}\right) = 0$
- The highest order derivative present is $\dfrac{d^4y}{dx^4}$.
Order: 4 - Since the equation is not polynomial in derivatives, because of $\sin\left(\dfrac{d^3y}{dx^3}\right)$, the degree is not defined.
These examples demonstrate how to identify order and degree from differential equations, a key concept in calculus and differential equations studies.
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NCERT Question.2.1 : Verify whether the given function
$$
y = a e^x + b e^{-x} + x^2
$$
is a solution to the differential equation
$$
x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} – xy + x^2 – 2 = 0
$$
Solution :
Given function :
$$
y = a e^x + b e^{-x} + x^2
$$
Step 1: Differentiate $y$ with respect to $x$
First derivative:
$$
\frac{dy}{dx} = a e^x – b e^{-x} + 2x
$$
Second derivative:
$$
\frac{d^2y}{dx^2} = a e^x + b e^{-x} + 2
$$
Step 2: Substitute into the differential equation
The left-hand side (LHS) is:
$$
LHS = x \frac{d^2y}{dx^2} + 2 \frac{dy}{dx} – xy + x^2 – 2
$$
Substituting the derivatives and $y$:
$$
LHS = x(a e^x + b e^{-x} + 2) + 2(a e^x – b e^{-x} + 2x) -\\- x(a e^x + b e^{-x} + x^2) + x^2 – 2
$$
Step 3: Simplify the expression
$$
LHS = (x a e^x + x b e^{-x} + 2x) + (2 a e^x – 2 b e^{-x} + 4x) – \\-(x a e^x + x b e^{-x} + x^3) + x^2 – 2
$$
$$
LHS = 2 a e^x – 2 b e^{-x} – x^3 + x^2 + 6x – 2
$$
Step 4: Compare LHS with RHS
The differential equation requires $LHS = 0$, but
$$
LHS = 2 a e^x – 2 b e^{-x} – x^3 + x^2 + 6x – 2 \neq 0
$$
Conclusion
Since the LHS is not equal to the RHS,
$$
y = a e^x + b e^{-x} + x^2
$$
is not a solution of the differential equation.
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NCERT Question.2.2 : Verify whether the given function
$$
y = e^x (a \cos x + b \sin x)
$$
is a solution to the differential equation
$$
\frac{d^2y}{dx^2} – 2 \frac{dy}{dx} + 2y = 0
$$
Solution :
Given function
$$
y = e^x (a \cos x + b \sin x)
$$
Step 1: Differentiate $y$ with respect to $x$
The given function can be expanded as:
$$
y = a e^x \cos x + b e^x \sin x
$$
First derivative:
$$
\frac{dy}{dx} = (a + b) e^x \cos x + (b – a) e^x \sin x \quad \text{… (1)}
$$
Second derivative:
$$
\frac{d^2y}{dx^2} = 2 e^x (b \cos x – a \sin x) \quad \text{… (2)}
$$
Step 2: Substitute into the differential equation
The left-hand side (LHS) is:
$$
LHS = \frac{d^2y}{dx^2} – 2 \frac{dy}{dx} + 2y
$$
Substituting (1) and (2):
$$
LHS = [2 e^x (b \cos x – a \sin x)] – 2[(a + b) e^x \cos x + (b – a) e^x \sin x] +\\+ 2 e^x (a \cos x + b \sin x)
$$
Step 3: Simplify the expression
$$
LHS = e^x \big[ (2b \cos x – 2a \sin x) – (2a \cos x + 2b \cos x + 2b \sin x – 2a \sin x) + \\+(2a \cos x + 2b \sin x) \big]
$$
$$
LHS = e^x [0] = 0
$$
Step 4: Compare LHS with RHS
Since
$$
LHS = 0 = RHS
$$
the given function satisfies the differential equation.
Conclusion
The function
$$
y = e^x (a \cos x + b \sin x)
$$
is a solution of the differential equation
$$
\frac{d^2y}{dx^2} – 2 \frac{dy}{dx} + 2y = 0
$$
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NCERT Question.2.3 : Verify whether the given function
$$
y = x \sin 3x
$$
is a solution to the differential equation
$$
\frac{d^2y}{dx^2} + 9y – 6 \cos 3x = 0
$$
Solution :
Given function
$$
y = x \sin 3x
$$
Step 1: Differentiate $y$ with respect to $x$
First derivative:
$$
\frac{dy}{dx} = \sin 3x + 3x \cos 3x \quad \text{… (1)}
$$
Second derivative:
$$
\frac{d^2y}{dx^2} = \frac{d}{dx} (\sin 3x) + 3 \frac{d}{dx} (x \cos 3x)
$$
$$
\frac{d^2y}{dx^2} = 3 \cos 3x + 3 (\cos 3x + x(-\sin 3x) \cdot 3)
$$
$$
\frac{d^2y}{dx^2} = 6 \cos 3x – 9x \sin 3x \quad \text{… (2)}
$$
Step 2: Substitute into the differential equation
The left-hand side (LHS) is:
$$
LHS = \frac{d^2y}{dx^2} + 9y – 6 \cos 3x
$$
Substituting (2) and $y = x \sin 3x$:
$$
LHS = (6 \cos 3x – 9x \sin 3x) + 9(x \sin 3x) – 6 \cos 3x
$$
Step 3: Simplify the expression
$$
LHS = 6 \cos 3x – 9x \sin 3x + 9x \sin 3x – 6 \cos 3x
$$
$$
LHS = 0
$$
Step 4: Compare LHS with RHS
Since
$$
LHS = 0 = RHS
$$
the given function satisfies the differential equation.
Conclusion
The function
$$
y = x \sin 3x
$$
is a solution of the differential equation
$$
\frac{d^2y}{dx^2} + 9y – 6 \cos 3x = 0
$$
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NCERT Question.2.4 : Verify whether the given function
$$
x^2 = 2y^2 \log y
$$
is a solution to the differential equation
$$
(x^2 + y^2) \frac{dy}{dx} – xy = 0
$$
Solution :
Given function
$$
x^2 = 2y^2 \log y
$$
Step 1: Differentiate $x^2 = 2y^2 \log y$ with respect to $x$
Differentiating both sides:
$$
2x = 2 \frac{d}{dx} (y^2 \log y)
$$
Simplifying the right-hand side:
$$
x = \frac{d}{dx} (y^2 \log y)
$$
Using the product and chain rules:
$$
x = 2y \log y \frac{dy}{dx} + y^2 \cdot \frac{1}{y} \frac{dy}{dx}
$$
$$
x = \frac{dy}{dx} \big[ 2y \log y + y \big]
$$
$$
\frac{dy}{dx} = \frac{x}{y(1 + 2 \log y)} \quad \text{… (1)}
$$
Step 2: Substitute into the differential equation
The left-hand side (LHS) is:
$$
LHS = (x^2 + y^2) \frac{dy}{dx} – xy
$$
Substituting (1) and $x^2 = 2y^2 \log y$:
$$
LHS = [2y^2 \log y + y^2] \cdot \frac{x}{y(1 + 2 \log y)} – xy
$$
$$
LHS = [y^2 (2 \log y + 1)] \cdot \frac{x}{y(1 + 2 \log y)} – xy
$$
$$
LHS = xy – xy
$$
$$
LHS = 0
$$
Step 3: Compare LHS with RHS
Since
$$
LHS = 0 = RHS
$$
the given function satisfies the differential equation.
Conclusion
The function
$$
x^2 = 2y^2 \log y
$$
is a solution of the differential equation
$$
(x^2 + y^2) \frac{dy}{dx} – xy = 0
$$
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NCERT Question 3: Prove that $x^2 – y^2 = C(x^2 + y^2)^2$ is the general solution of the differential equation
$$(x^3 – 3xy^2)dx = (y^3 – 3x^2y)dy$$
where $C$ is a parameter.
Solution:
Given differential equation:
$$(x^3 – 3xy^2)dx = (y^3 – 3x^2y)dy$$
Rewriting,
$$\frac{dy}{dx} = \frac{x^3 – 3xy^2}{y^3 – 3x^2y} \quad …(1)$$
Let $y = vx$, then
$$\frac{dy}{dx} = v + x\frac{dv}{dx} \quad …(2)$$
Substituting in (1):
$$v + x\frac{dv}{dx} = \frac{x^3 – 3x(vx)^2}{(vx)^3 – 3x^2(vx)}$$
$$v + x\frac{dv}{dx} = \frac{1 – 3v^2}{v^3 – 3v}$$
Hence,
$$x\frac{dv}{dx} = \frac{1 – 3v^2}{v^3 – 3v} – v$$
$$x\frac{dv}{dx} = \frac{1 – v^4}{v^3 – 3v}$$
Rearranging,
$$\frac{v^3 – 3v}{1 – v^4}dv = \frac{dx}{x}$$
Integrating both sides,
$$\int \frac{v^3 – 3v}{1 – v^4}dv = \log x + \log C’ \quad …(3)$$
Splitting the integral,
$$\int \frac{v^3 – 3v}{1 – v^4}dv = \int \frac{v^3}{1 – v^4}dv – 3\int \frac{v}{1 – v^4}dv$$
Let
$$I_1 = \int \frac{v^3}{1 – v^4}dv, \quad I_2 = \int \frac{v}{1 – v^4}dv$$
Now, for $I_1$: let $t = 1 – v^4$, then $dt = -4v^3dv$,
$$I_1 = \int -\frac{1}{4}\frac{dt}{t} = -\frac{1}{4}\log(1 – v^4) \quad …(4)$$
For $I_2$: let $p = v^2$, then $dp = 2vdv$,
$$I_2 = \frac{1}{2}\int \frac{dp}{1 – p^2} = \frac{1}{4}\log\left|\frac{1 + p}{1 – p}\right|$$
$$I_2 = \frac{1}{4}\log\left|\frac{1 + v^2}{1 – v^2}\right| \quad …(5)$$
Substituting (4) and (5) in (3):
$$\int \frac{v^3 – 3v}{1 – v^4}dv = -\frac{1}{4}\log(1 – v^4) – \frac{3}{4}\log\left|\frac{1 + v^2}{1 – v^2}\right|$$
Therefore,
$$-\frac{1}{4}\log(1 – v^4) – \frac{3}{4}\log\left|\frac{1 + v^2}{1 – v^2}\right| = \log(C’x)$$
Combining logarithms,
$$-\frac{1}{4}\log\left[(1 – v^4)\left|\frac{1 + v^2}{1 – v^2}\right|^3\right] = \log(C’x)$$
Simplifying,
$$(1 – v^4)\left|\frac{1 + v^2}{1 – v^2}\right|^3 = (C’x)^{-4}$$
Since $1 – v^4 = (1 – v^2)(1 + v^2)$,
$$\frac{(1 + v^2)^4}{(1 – v^2)^2} = (C’x)^{-4}$$
Replacing $v = \dfrac{y}{x}$,
$$\frac{\left(1 + \dfrac{y^2}{x^2}\right)^4}{\left(1 – \dfrac{y^2}{x^2}\right)^2} = (C’x)^{-4}$$
Multiplying by $x^8$,
$$\frac{(x^2 + y^2)^4}{(x^2 – y^2)^2} = (C’x)^{-4}$$
Simplifying,
$$(x^2 – y^2)^2 = (C’x)^4(x^2 + y^2)^4$$
Taking square roots,
$$x^2 – y^2 = C'(x^2 + y^2)^2$$
Let $C = C’$, then
$$x^2 – y^2 = C(x^2 + y^2)^2$$
Hence proved that the general solution of the differential equation
$$(x^3 – 3xy^2)dx = (y^3 – 3x^2y)dy$$
is
$$x^2 – y^2 = C(x^2 + y^2)^2.$$
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NCERT Question.4 : Find the general solution of the differential equation
$$\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0.$$
Solution:
Given
$$\frac{dy}{dx} + \sqrt{\frac{1-y^2}{1-x^2}} = 0.$$
So
$$\frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}.$$
Separate variables:
$$\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}.$$
Integrate both sides:
$$\int \frac{dy}{\sqrt{1-y^2}} = -\int \frac{dx}{\sqrt{1-x^2}}.$$
Thus
$$\sin^{-1}y = -\sin^{-1}x + C.$$
Rearrange to obtain the general solution:
$$\sin^{-1}x + \sin^{-1}y = C.$$
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NCERT Question.5 : Show that the general solution of the differential equation
$$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$$
is given by
$$(x + y + 1) = A(1 – x – y – 2xy),$$
where $A$ is the parameter.
Solution:
Given differential equation:
$$\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0$$
Rearranging, we get
$$\frac{dy}{dx} = -\frac{y^2 + y + 1}{x^2 + x + 1}$$
So,
$$\frac{dy}{y^2 + y + 1} = -\frac{dx}{x^2 + x + 1}$$
Integrate both sides:
$$\int \frac{dy}{y^2 + y + 1} = -\int \frac{dx}{x^2 + x + 1}$$
We can write
$$y^2 + y + 1 = \left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$
and
$$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$
Hence,
$$\int \frac{dy}{(y + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = -\int \frac{dx}{(x + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$$
Using the formula $\int \dfrac{du}{u^2 + a^2} = \dfrac{1}{a}\tan^{-1}\left(\dfrac{u}{a}\right)$, we get
$$\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = -\frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C$$
Simplifying,
$$\tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right) + \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) = \frac{\sqrt{3}C}{2}$$
Using the tangent addition formula,
$$\dfrac{\dfrac{2x + 2y + 2}{\sqrt{3}}}{1 – \dfrac{(2x + 1)(2y + 1)}{3}} = \tan\left(\dfrac{\sqrt{3}C}{2}\right)$$
Simplify further:
$$\dfrac{\dfrac{2x + 2y + 2}{\sqrt{3}}}{\dfrac{3 – (4xy + 2x + 2y + 1)}{3}} = C_1, \quad \text{where } C_1 = \tan\left(\dfrac{\sqrt{3}C}{2}\right)$$
Hence,
$$\frac{\sqrt{3}(2x + 2y + 2)}{3 – (4xy + 2x + 2y + 1)} = C_1$$
Simplify the numerator and denominator:
$$2\sqrt{3}(x + y + 1) = C_1(2 – 4xy – 2x – 2y)$$
Dividing both sides by $2$,
$$\sqrt{3}(x + y + 1) = C_1(1 – x – y – 2xy)$$
Let
$$A = \frac{C_1}{\sqrt{3}}$$
Thus,
$$(x + y + 1) = A(1 – x – y – 2xy)$$
Hence proved. ✅
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NCERT Question.6 : Form the differential equation representing the family of curves given by $ (x- a)^2 + 2y^2 = a^2 $ where $ a $ is an arbitrary constant.
Solution :
Given equation:
$$(x – a)^2 + 2y^2 = a^2$$
The given equation can be written as:
$$x^2 + a^2 – 2ax + 2y^2 = a^2$$
On rearranging the above equation, we get
$$2y^2 = 2ax – x^2 \qquad\text{(1)}$$
Now, differentiate equation (1) with respect to $x$,
$$2\cdot 2y\frac{dy}{dx} = 2a – 2x$$
$$2y\frac{dy}{dx} = \frac{2a – 2x}{2}$$
$$\frac{dy}{dx} = \frac{a – x}{2y}$$
$$\frac{dy}{dx} = \frac{2ax – 2x^2}{4xy} \qquad\text{(2)}$$
From equation (1), we get
$$2ax = 2y^2 + x^2$$
Substitute the value in equation (2), and we get
$$\frac{dy}{dx} = \frac{2y^2 + x^2 – 2x^2}{4xy}$$
$$\frac{dy}{dx} = \frac{2y^2 – x^2}{4xy}$$
Therefore, the differential equation representing the family of curves given by $(x- a)^2 + 2y^2 = a^2$ is
$$\frac{dy}{dx} = \frac{2y^2 – x^2}{4xy}.$$
