Anand Classes presents a free PDF download of NCERT Solutions for Class 12 Maths Chapter 9 – Differential Equations, Exercise 9.4 (Set-2), designed especially for those looking to deepen their understanding of homogeneous differential equations and separable variable methods. These solutions follow the CBSE/NCERT syllabus precisely, include detailed step-by-step workings for general and particular solutions, and are perfect for Class 12 board exam preparation and competitive revision. Click the print button to download study material and notes.
NCERT Question 7 : Show that the given differential equation is homogeneous and solve it.
$$\left[x\cos\left(\frac{y}{x}\right) + y\sin\left(\frac{y}{x}\right)\right]ydx = \left[y\sin\left(\frac{y}{x}\right) – x\cos\left(\frac{y}{x}\right)\right]xdy$$
Solution :
Given differential equation
$$\left[x\cos\left(\frac{y}{x}\right) + y\sin\left(\frac{y}{x}\right)\right]ydx = \left[y\sin\left(\frac{y}{x}\right) – x\cos\left(\frac{y}{x}\right)\right]xdy$$
Rearranging,
$$\dfrac{dy}{dx} = \dfrac{\left[x\cos\left(\dfrac{y}{x}\right) + y\sin\left(\dfrac{y}{x}\right)\right]y}{\left[y\sin\left(\dfrac{y}{x}\right) – x\cos\left(\dfrac{y}{x}\right)\right]x}.$$
Since replacing $(x, y)$ with $(\lambda x, \lambda y)$ leaves the expression unchanged, the equation is homogeneous.
Let $y = vx$ so that $v = \dfrac{y}{x}$ and
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substitute $y = vx$ into the right-hand side:
$$
\frac{dy}{dx} = \frac{[x\cos v + vx\sin v]vx}{[vx\sin v – x\cos v]x}
= \frac{v(\cos v + v\sin v)}{v\sin v – \cos v}.
$$
Equating both sides,
$$
v + x\frac{dv}{dx} = \frac{v(\cos v + v\sin v)}{v\sin v – \cos v}.
$$
Simplify:
$$
x\frac{dv}{dx} = \frac{v(\cos v + v\sin v)}{v\sin v – \cos v} – v.
$$
Taking the LCM,
$$
x\frac{dv}{dx} = \frac{2v\cos v}{v\sin v – \cos v}.
$$
Separate the variables:
$$
\frac{v\sin v – \cos v}{2v\cos v}dv = \frac{dx}{x}.
$$
Simplify the integrand:
$$
\frac{v\sin v – \cos v}{2v\cos v} = \frac{1}{2}\left(\tan v – \frac{1}{v}\right).
$$
Integrating both sides,
$$
\int \frac{1}{2}\left(\tan v – \frac{1}{v}\right)dv = \int \frac{dx}{x}.
$$
Compute the integrals:
$$
\frac{1}{2}(-\ln|\cos v| – \ln|v|) = \ln|x| + C.
$$
Simplify:
$$
-\frac{1}{2}\ln(v\cos v) = \ln x + C.
$$
Rearranging,
$$
\ln(v\cos v) = -2\ln x + C’.
$$
Exponentiating both sides:
$$
v\cos v = \frac{K}{x^2},
$$
where $K = e^{C’}$.
Substitute back $v = \dfrac{y}{x}$:
$$
\frac{y}{x}\cos\left(\frac{y}{x}\right) = \frac{K}{x^2}.
$$
Multiply both sides by $x^2$:
$$
xy\cos\left(\frac{y}{x}\right) = K.
$$
Final Answer
$$\boxed{xy\cos\left(\frac{y}{x}\right) = C}$$
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NCERT Question 8 : Show that the given differential equation is homogeneous and solve it.
$$x\frac{dy}{dx} – y + x\sin\left(\frac{y}{x}\right) = 0.$$
Solution :
Given differential equation
$$x\frac{dy}{dx} – y + x\sin\left(\frac{y}{x}\right) = 0.$$
Rearranging, we get
$$x\frac{dy}{dx} = y – x\sin\left(\frac{y}{x}\right)$$
which gives
$$\frac{dy}{dx} = \frac{y – x\sin\left(\dfrac{y}{x}\right)}{x}.$$
Since the right-hand side is a function of $\dfrac{y}{x}$ only, the given equation is homogeneous.
Let $y = vx$, so that $v = \frac{y}{x}$ and
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substitute $y = vx$ into the equation:
$$
v + x\frac{dv}{dx} = \frac{vx – x\sin v}{x} = v – \sin v.
$$
Simplify:
$$
x\frac{dv}{dx} = -\sin v.
$$
Separate the variables:
$$
\frac{dv}{\sin v} = -\frac{dx}{x}.
$$
Integrate both sides:
$$
\int \csc vdv = -\int \frac{dx}{x}.
$$
Using $\displaystyle \int \csc vdv = \ln|\csc v – \cot v| + C$, we get
$$
\ln|\csc v – \cot v| = -\ln|x| + C.
$$
Simplify:
$$
\ln|\csc v – \cot v| = \ln\left(\frac{C}{x}\right).
$$
Exponentiating both sides:
$$
\csc v – \cot v = \frac{C}{x}.
$$
Substitute back $v = \dfrac{y}{x}$:
$$
\csc\left(\frac{y}{x}\right) – \cot\left(\frac{y}{x}\right) = \frac{C}{x}.
$$
We can also write it as:
$$
\frac{1 – \cos\left(\dfrac{y}{x}\right)}{\sin\left(\dfrac{y}{x}\right)} = \frac{C}{x}
$$
which gives
$$
x\left[1 – \cos\left(\frac{y}{x}\right)\right] = C\sin\left(\frac{y}{x}\right).
$$
Final Answer
$$\boxed{\csc\left(\frac{y}{x}\right) – \cot\left(\frac{y}{x}\right) = \frac{C}{x}}$$
or equivalently
$$\boxed{x\left[1 – \cos\left(\frac{y}{x}\right)\right] = C\sin\left(\frac{y}{x}\right)}.$$
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NCERT Question 9 : Show that the given differential equation is homogeneous and solve it.
$$ydx + x\log\left(\frac{y}{x}\right)dy – 2xdy = 0$$
Solution :
Given differential equation
$$ydx + x\log\left(\frac{y}{x}\right)dy – 2xdy = 0$$
Rewriting,
$$ydx + x\left[\log\left(\dfrac{y}{x}\right) – 2\right]dy = 0.$$
Hence,
$$\frac{dy}{dx} = \frac{y}{x\left(2 – \log\left(\dfrac{y}{x}\right)\right)}.$$
The right-hand side is a function of $\dfrac{y}{x}$ only, so the equation is homogeneous.
Let $y = vx$, so that $v = \frac{y}{x}$ and
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substitute $y = vx$ into the differential equation:
$$
v + x\frac{dv}{dx} = \frac{v}{2 – \log v}.
$$
Simplify:
$$
x\frac{dv}{dx} = \frac{v}{2 – \log v} – v = \frac{v(\log v – 1)}{2 – \log v}.
$$
Separate the variables:
$$
\frac{2 – \log v}{v(\log v – 1)}dv = \frac{dx}{x}.
$$
Let $\log v = t$ so that $dt = \dfrac{dv}{v}$. Then,
$$
\frac{2 – t}{t – 1}dt = \frac{dx}{x}.
$$
Simplify the left-hand side:
$$
\frac{2 – t}{t – 1} = -1 + \frac{1}{t – 1}.
$$
Integrate both sides:
$$
\int \left(-1 + \frac{1}{t – 1}\right)dt = \int \frac{dx}{x}.
$$
On integration,
$$
-t + \log|t – 1| = \log|x| + C.
$$
Substitute back $t = \log v$:
$$
-\log v + \log|\log v – 1| = \log|x| + C.
$$
Combine the logarithms:
$$
\log\left(\frac{\log v – 1}{v}\right) = \log|x| + C.
$$
Simplify:
$$
\frac{\log v – 1}{v} = C_1 x
$$
where $C_1 = e^{C}$.
Substitute back $v = \dfrac{y}{x}$:
$$
\frac{\log\left(\dfrac{y}{x}\right) – 1}{\dfrac{y}{x}} = C_1 x.
$$
Simplify:
$$
\frac{\log\left(\dfrac{y}{x}\right) – 1}{y} = C_1.
$$
Hence, the general solution is
$$
\boxed{\log\left(\frac{y}{x}\right) – 1 = C y.}
$$
Final Answer
$$\boxed{\log\left(\frac{y}{x}\right) – 1 = C y}$$
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NCERT Question.10 : Show that the given differential equation is homogeneous and solve it.
$$\left[1+e^{\dfrac{x}{y}}\right]dx + e^{\dfrac{x}{y}}\left(1-\dfrac{x}{y}\right)dy = 0$$
Solution
Given differential equation
$$\left[1+e^{\dfrac{x}{y}}\right]dx + e^{\dfrac{x}{y}}\left(1-\dfrac{x}{y}\right)dy = 0$$
Rearrange to obtain
$$\frac{dx}{dy} = -\frac{e^{\dfrac{x}{y}}\left(1-\dfrac{x}{y}\right)}{1+e^{\dfrac{x}{y}}}$$
This equation is homogeneous (depends on $x/y$). Let
$$v = \frac{x}{y}, \quad x = vy, \quad \frac{dx}{dy} = v + y\frac{dv}{dy}$$
Substitute in the equation:
$$v + y\frac{dv}{dy} = -\frac{e^v(1-v)}{1+e^v}$$
Simplify:
$$y\frac{dv}{dy} = -\frac{e^v(1-v) + v(1+e^v)}{1+e^v} = -\frac{v+e^v}{1+e^v}$$
Separate the variables:
$$\frac{1+e^v}{v+e^v}dv = -\frac{dy}{y}$$
Since
$$\frac{d}{dv}\ln(v+e^v) = \frac{1+e^v}{v+e^v}$$
Integrate both sides:
$$\ln(v+e^v) = -\ln|y| + C$$
Exponentiate and substitute $v = \dfrac{x}{y}$:
$$v + e^v = \frac{C_1}{y} \Rightarrow \frac{x}{y} + e^{\dfrac{x}{y}} = \frac{C_1}{y}$$
Multiply through by $y$:
$$x + y e^{\dfrac{x}{y}} = C$$
Final Answer
$$\boxed{x + y e^{\dfrac{x}{y}} = C}$$
NCERT Question 11 : Find the particular solution satisfying $y = 1$ when $x = 1$ for the differential equation
$$(x + y)dy + (x – y)dx = 0.$$
Solution :
Given Equation
$$(x + y)dy + (x – y)dx = 0.$$
Rearranging, we get
$$\frac{dy}{dx} = -\frac{x – y}{x + y}.$$
This equation is homogeneous. Let $y = vx$, where $v = \dfrac{y}{x}$ and hence $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
Substitute into the equation:
$$v + x\frac{dv}{dx} = -\frac{1 – v}{1 + v}.$$
Simplifying,
$$x\frac{dv}{dx} = -\frac{1 – v}{1 + v} – v = -\frac{1 – v + v(1 + v)}{1 + v} = -\frac{1 + v^2}{1 + v}.$$
Separating the variables,
$$\frac{1 + v}{1 + v^2}dv = -\frac{dx}{x}.$$
Integrating both sides,
$$\int \frac{1 + v}{1 + v^2}dv = -\int \frac{dx}{x}.$$
We can split the integral:
$$\int \frac{1 + v}{1 + v^2}dv = \int \frac{dv}{1 + v^2} + \int \frac{vdv}{1 + v^2}.$$
Thus,
$$tan^{-1}v + \frac{1}{2}\ln(1 + v^2) = -\ln|x| + C.$$
Rewriting,
$$\ln|x| + \frac{1}{2}\ln(1 + v^2) + tan^{-1}v = C.$$
Substitute back $v = \dfrac{y}{x}$:
$$\frac{1}{2}\ln(x^2 + y^2) + tan^{-1}\left(\frac{y}{x}\right) = C.$$
Now, use the condition $y = 1$ when $x = 1$:
$$\frac{1}{2}\ln(1^2 + 1^2) + tan^{-1}(1) = \frac{1}{2}\ln 2 + \frac{\pi}{4} = C.$$
Hence, the particular solution is
$$\boxed{\frac{1}{2}\ln(x^2 + y^2) + tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2}\ln 2 + \frac{\pi}{4}}$$
or equivalently,
$$\boxed{\ln(x^2 + y^2) + 2tan^{-1}\left(\frac{y}{x}\right) = \ln 2 + \frac{\pi}{2}}.$$
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