Anand Classes provides a detailed and free downloadable PDF of NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations, Exercise 9.4 (Set-3) designed to help students master the concepts of homogeneous differential equations through clear, step-by-step solutions. This set includes advanced problems that involve substitution methods such as y=vx, separation of variables, and integration techniques that build a solid understanding of calculus fundamentals. The solutions strictly follow the latest CBSE and NCERT syllabus, making them ideal for both board exam preparation and competitive exams like JEE Main. Each question is explained systematically by expert teachers at Anand Classes to enhance problem-solving accuracy and conceptual clarity. Click the print button to download study material and notes.
NCERT Question 12 : Find the particular solution satisfying $y = 1$ when $x = 1$ for the differential equation
$$x^2dy + (xy + y^2)dx = 0.$$
Solution :
Given differential equation
$$x^2dy + (xy + y^2)dx = 0.$$
Rearranging, we get
$$\frac{dy}{dx} = -\frac{xy + y^2}{x^2} = -\left(\frac{y}{x} + \frac{y^2}{x^2}\right).$$
Let $y = vx$, where $v = \dfrac{y}{x}$ and hence $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
Substitute this in:
$$v + x\frac{dv}{dx} = -\left(v + v^2\right).$$
Simplifying,
$$x\frac{dv}{dx} = -v(v + 2).$$
Separate variables:
$$\frac{dv}{v(v + 2)} = -\frac{dx}{x}.$$
Now, resolve into partial fractions:
$$\frac{1}{v(v + 2)} = \frac{A}{v} + \frac{B}{v + 2}.$$
So,
$$1 = A(v + 2) + Bv \Rightarrow (A + B)v + 2A = 1.$$
Comparing coefficients:
$$A + B = 0, \quad 2A = 1 \Rightarrow A = \frac{1}{2}, , B = -\frac{1}{2}.$$
Thus,
$$\frac{dv}{v(v + 2)} = \frac{1}{2}\left(\frac{1}{v} – \frac{1}{v + 2}\right)dv.$$
Integrating both sides:
$$\frac{1}{2}\left(\ln|v| – \ln|v + 2|\right) = -\ln|x| + C.$$
Simplify:
$$\ln\left|\frac{v}{v + 2}\right| = -2\ln|x| + C’.$$
Taking exponentials:
$$\frac{v}{v + 2} = \frac{K}{x^2}.$$
Substitute $v = \dfrac{y}{x}$:
$$\dfrac{\dfrac{y}{x}}{\dfrac{y}{x} + 2} = \dfrac{K}{x^2}.$$
Simplify:
$$\frac{y}{y + 2x} = \frac{K}{x^2}.$$
Rearranging:
$$y + 2x = \frac{x^2y}{K}.$$
Now, using the condition $y = 1$ when $x = 1$:
$$1 + 2(1) = \frac{1^2(1)}{K} \Rightarrow 3 = \frac{1}{K} \Rightarrow K = \frac{1}{3}.$$
Substitute back:
$$y + 2x = 3x^2y.$$
Hence, the required particular solution is
$$\boxed{y + 2x = 3x^2y.}$$
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NCERT Question.13 : Solve the differential equation
$$\left[x\sin^2\left(\dfrac{y}{x}\right) – y\right]dx + xdy = 0$$
given that $y = \dfrac{\pi}{4}$ when $x = 1.$
Solution :
Given differential equation :
$$\left[x\sin^2\left(\dfrac{y}{x}\right) – y\right]dx + xdy = 0$$
Divide through by $xdx$:
$$\frac{dy}{dx} = \frac{y}{x} – \sin^2\left(\frac{y}{x}\right).$$
Let $y = vx$, so that
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substitute into the equation:
$$v + x\frac{dv}{dx} = v – \sin^2 v.$$
Simplify:
$$x\frac{dv}{dx} = -\sin^2 v.$$
Separate the variables:
$$\frac{dv}{\sin^2 v} = -\frac{dx}{x}.$$
Integrate both sides:
$$\int \csc^2 vdv = -\int \frac{dx}{x}.$$
We get:
$$-\cot v = -\ln|x| + C$$
or
$$\cot v = \ln|x| + C.$$
Substitute back $v = \dfrac{y}{x}$:
$$\cot\left(\frac{y}{x}\right) = \ln|x| + C.$$
Applying Initial Condition
Given $y = \dfrac{\pi}{4}$ when $x = 1$:
$$\cot\left(\frac{\pi}{4}\right) = \ln(1) + C.$$
This gives $1 = 0 + C$, so $C = 1.$
Final Answer
$$\boxed{\cot\left(\frac{y}{x}\right) = 1 + \ln|x|}$$
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NCERT Question.14 : Find the particular solution satisfying $y = 0$ when $x = 1$ for the differential equation
$$\frac{dy}{dx} – \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0, \qquad y(1) = 0.$$
Solution :
We have the differential equation
$$\frac{dy}{dx} – \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0$$
Let $v = \dfrac{y}{x}$ so $y = vx$ and
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substituting into the given equation:
$$v + x\frac{dv}{dx} – v + \csc v = 0 \quad \Rightarrow \quad x\frac{dv}{dx} = -\csc v.$$
Separating variables:
$$\sin vdv = -\frac{dx}{x}.$$
Integrating both sides:
$$\int \sin vdv = -\int \frac{dx}{x} \quad \Rightarrow \quad -\cos v = -\ln|x| + C.$$
Hence,
$$\cos v = \ln|x| + C.$$
Substituting back $v = \dfrac{y}{x}$ gives
$$\boxed{\cos\left(\frac{y}{x}\right) = \ln|x| + C.}$$
Applying the initial condition $y(1) = 0$ (i.e., $v(1) = 0$), we have $\cos 0 = 1$ and $\ln 1 = 0$, so $1 = 0 + C$ or $C = 1$.
Therefore, the particular solution is
$$\boxed{\cos\left(\frac{y}{x}\right) = 1 + \ln|x|}$$
$$\boxed{\cos\left(\frac{y}{x}\right) = \ln e + \ln|x|}$$
$$\boxed{\cos\left(\frac{y}{x}\right) = \ln|ex|}$$
or equivalently,
$$\boxed{\frac{y}{x} = \arccos(1 + \ln|x|)}.$$
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NCERT Question 15 : Find the particular solution satisfying $y = 2$ when $x = 1$ for the differential equation
$$2xy + y^2 – 2x^2\frac{dy}{dx} = 0,\qquad y(1)=2.$$
Solution:
Given differential equation
$$2xy + y^2 – 2x^2\frac{dy}{dx} = 0$$
Rearrange
$$2x^2\frac{dy}{dx} = 2xy + y^2$$
$$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}.$$
Put $y = vx$ so $v=\dfrac{y}{x}$ and $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$. Substitute
$$v + x\frac{dv}{dx} = \frac{2v + v^2}{2} = v + \frac{v^2}{2}.$$
Thus
$$x\frac{dv}{dx} = \frac{v^2}{2}.$$
Separate variables
$$\frac{dv}{v^2} = \frac{dx}{2x}.$$
Integrate
$$\int v^{-2}dv = \int \frac{dx}{2x}$$
$$-\frac{1}{v} = \frac{1}{2}\ln|x| + C.$$
So
$$\frac{1}{v} = -\frac{1}{2}\ln|x| – C$$
$$v = -\frac{1}{\tfrac{1}{2}\ln|x| + C’}$$
where $C’ = C$.
Use $y(1)=2$ so $v(1)=2$ and $\ln 1=0$ therefore
$$2 = -\frac{1}{C’}\quad\Rightarrow\quad C’ = -\tfrac{1}{2}.$$
Hence
$$v = -\frac{1}{\tfrac{1}{2}\ln|x| – \tfrac{1}{2}} = \frac{2}{1 – \ln|x|}.$$
Finally
$$\boxed{y = \frac{2x}{1 – \ln|x|}}.$$
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NCERT Question 16 : A homogeneous differential equation of the form
$$\frac{dx}{dy} = h\left(\frac{x}{y}\right)$$
can be solved by making which substitution?
Options:
(A) $y = vx$
(B) $v = yx$
(C) $x = vy$
(D) $x = v$
Solution :
The given differential equation is
$$\frac{dx}{dy} = h\left(\frac{x}{y}\right).$$
Here, the function on the right-hand side depends on the ratio $\dfrac{x}{y}$.
To simplify such equations, we substitute
$$x = vy,$$
which means
$$v = \frac{x}{y}.$$
Now, differentiating $x = vy$ with respect to $y$:
$$\frac{dx}{dy} = v + y\frac{dv}{dy}.$$
Substitute this in the given equation:
$$v + y\frac{dv}{dy} = h(v).$$
Rearranging gives:
$$y\frac{dv}{dy} = h(v) – v.$$
This is a separable differential equation in $v$ and $y$, which can be solved by integration.
Hence, the correct substitution is
$$\boxed{x = vy.}$$
Correct Option:
(C) $x = vy$
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NCERT Question 17 : Which of the following is a homogeneous differential equation?
Options:
(A) $(4x + 6y + 5)dy – (3y + 2x + 4)dx = 0$
(B) $(xy)dx – (x^3 + y^3)dy = 0$
(C) $(x^3 + 2y^2)dx + 2xydy = 0$
(D) $y^2dx + (x^2 – xy – y^2)dy = 0$
Solution :
A differential equation is homogeneous if each term of the equation is of the same degree in $x$ and $y$.
Let’s check each option one by one.
Option (A)
$(4x + 6y + 5)dy – (3y + 2x + 4)dx = 0$
- Terms like $4x$ and $6y$ are of degree 1.
- The constants $5$ and $4$ are of degree 0.
Since all terms are not of the same degree, this is not homogeneous.
Option (B)
$(xy)dx – (x^3 + y^3)dy = 0$
- The term $xy$ is of degree 2 (because $x^1y^1$).
- The terms $x^3$ and $y^3$ are of degree 3.
Since degrees are not equal, this is not homogeneous.
Option (C)
$(x^3 + 2y^2)dx + 2xydy = 0$
- $x^3$ is of degree 3, but $2y^2$ is of degree 2.
- $2xy$ is of degree 2.
Again, all terms are not of the same degree, so this is not homogeneous.
Option (D)
$y^2dx + (x^2 – xy – y^2)dy = 0$
- In the first term, $y^2$ has degree 2.
- In the second term, each of $x^2$, $xy$, and $y^2$ has degree 2.
All terms are of degree 2, hence the equation is homogeneous.
Correct Option:
(D) $y^2dx + (x^2 – xy – y^2)dy = 0$
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