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NCERT Question 1 : Show that the differential equation $(x^{2} + xy)dy = (x^{2} + y^{2})dx$ is homogeneous, and solve it.
Solution :
Checking Homogeneity
The given equation can be written as
$$
\frac{dy}{dx} = \frac{x^{2} + y^{2}}{x^{2} + xy}
$$
Both numerator and denominator are homogeneous functions of degree $2$.
For any scalar $\lambda$,
$$
\frac{(\lambda x)^{2} + (\lambda y)^{2}}{(\lambda x)^{2} + (\lambda x)(\lambda y)}
= \frac{\lambda^{2}(x^{2} + y^{2})}{\lambda^{2}(x^{2} + xy)}
= \frac{x^{2} + y^{2}}{x^{2} + xy}
$$
Hence, the given differential equation is homogeneous.
Let $y = vx$, then
$$
\frac{dy}{dx} = v + x\frac{dv}{dx}
$$
Substitute this into the given equation:
$$
v + x\frac{dv}{dx} = \frac{x^{2} + v^{2}x^{2}}{x^{2} + vx^{2}} = \frac{1 + v^{2}}{1 + v}
$$
Simplify:
$$
x\frac{dv}{dx} = \frac{1 + v^{2}}{1 + v} – v = \frac{1 – v}{1 + v}
$$
Separate the variables:
$$
\frac{1 + v}{1 – v}dv = \frac{dx}{x}
$$
Integration
Rewrite $\dfrac{1 + v}{1 – v}$ as
$$
\frac{1 + v}{1 – v} = -1 + \frac{2}{1 – v}
$$
Hence,
$$
\int \frac{1 + v}{1 – v}dv = \int \left(-1 + \frac{2}{1 – v}\right)dv = -v – 2\ln|1 – v|
$$
Therefore,
$$
-v – 2\ln|1 – v| = \ln|x| + C
$$
or
$$
v + 2\ln|1 – v| = -\ln|x| + C_1
$$
Simplify:
$$
\ln\big((1 – v)^2\big) + v = \ln\left(\dfrac{C_2}{x}\right)
$$
Exponentiating
$$
(1 – v)^2 e^{v} = \dfrac{C_2}{x}
$$
Substitute $v = \dfrac{y}{x}$:
$$
\left(1 – \dfrac{y}{x}\right)^2 e^{y/x} = \dfrac{C_2}{x}
$$
Multiply both sides by $x$:
$$
(x – y)^2 e^{y/x} = Cx
$$
Final Answer
$$
\boxed{(x – y)^2 e^{\dfrac{y}{x}} = Cx}
$$
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NCERT Question 2 : Show that the differential equation $$y’=\dfrac{x+y}{x}$$ is homogeneous, and solve it.
Solution :
Given Equation :
$$\dfrac{dy}{dx} =\dfrac{x+y}{x}$$
Rewrite the right-hand side:
$$\frac{x+y}{x}=1+\frac{y}{x}.$$
This is homogeneous of degree $0$. Use the substitution $y=vx$ (so $v=\dfrac{y}{x}$). Then
$$\frac{dy}{dx}=v+x\frac{dv}{dx}.$$
Substitute into the equation:
$$v+x\frac{dv}{dx}=1+v\quad\Longrightarrow\quad x\frac{dv}{dx}=1.$$
Separate and integrate:
$$dv=\frac{dx}{x}\quad\Longrightarrow\quad v=\ln|x|+C.$$
Substitute back $v=\dfrac{y}{x}$:
$$\frac{y}{x}=\ln|x|+C.$$
Final answer
$$\boxed{y=x\ln|x|+Cx}\qquad C\ \text{an arbitrary constant}$$
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NCERT Question 3 : Show that the differential equation
$$(x – y)dy – (x + y)dx = 0$$
is homogeneous, and solve it.
Solution :
Check for Homogeneity
The given equation can be written as
$$\frac{dy}{dx} = \frac{x + y}{x – y}.$$
Here, both numerator $(x + y)$ and denominator $(x – y)$ are homogeneous functions of degree $1$.
Therefore, the differential equation is homogeneous of degree 0.
Substitution
Let $y = vx$, so that
$$\frac{dy}{dx} = v + x\frac{dv}{dx}.$$
Substitute these values into the equation:
$$v + x\frac{dv}{dx} = \frac{x + vx}{x – vx} = \frac{1 + v}{1 – v}.$$
Simplifying,
$$x\frac{dv}{dx} = \frac{1 + v}{1 – v} – v = \frac{1 + v^2}{1 – v}.$$
Separation of Variables
$$\frac{1 – v}{1 + v^2}dv = \frac{dx}{x}.$$
Integrate both sides:
$$\int \frac{1 – v}{1 + v^2}dv = \int \frac{dx}{x}.$$
Now,
$$\int \frac{1 – v}{1 + v^2}dv = \int \frac{dv}{1 + v^2} – \int \frac{vdv}{1 + v^2} = \tan^{-1}v – \frac{1}{2}\ln(1 + v^2).$$
Thus,
$$\tan^{-1}v – \frac{1}{2}\ln(1 + v^2) = \ln|x| + C.$$
Back Substitution
Since $v = \dfrac{y}{x}$,
$$\tan^{-1}\left(\frac{y}{x}\right) – \frac{1}{2}\ln\left(1 + \frac{y^2}{x^2}\right) = \ln|x| + C.$$
Simplify the logarithmic term:
$$\tan^{-1}\left(\frac{y}{x}\right) – \frac{1}{2}\ln(x^2 + y^2) = C.$$
Final Answer
$$\boxed{\tan^{-1}\left(\frac{y}{x}\right) – \frac{1}{2}\ln(x^2 + y^2) = C}$$
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NCERT Question 4 : Show that the differential equation
$$ (x^{2}-y^{2})dx + 2xydy = 0. $$
is homogeneous, and solve it.
Solution :
Check & substitution
Both $M(x, y) = x^{2} – y^{2}$ and $N(x, y) = 2xy$ are homogeneous functions of degree 2.
Let $y = vx$, so that $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$.
Substituting $y = vx$ into the given equation, we get
$$ (x^{2} – v^{2}x^{2})dx + 2x(vx)d(vx) = 0 $$
Expanding,
$$ x^{2}(1 – v^{2})dx + 2v x^{2}(vdx + xdv) = 0 $$
Simplifying,
$$ x^{2}(1 – v^{2})dx + 2v^{2}x^{2}dx + 2v x^{3}dv = 0 $$
$$ x^{2}(1 + v^{2})dx + 2v x^{3}dv = 0 $$
Separation of Variables
Divide throughout by $x^{2}$ (assuming $x \neq 0$ ):
$$ (1 + v^{2})dx + 2v xdv = 0 $$
Rearranging,
$$ \frac{dx}{x} = -\frac{2v}{1 + v^{2}}dv $$
Integration
Integrating both sides,
$$ \int \frac{dx}{x} = -\int \frac{2v}{1 + v^{2}}dv $$
$$ \ln|x| = -\ln(1 + v^{2}) + C $$
Combine the logarithms:
$$ \ln|x(1 + v^{2})| = C $$
Taking the exponential on both sides,
$$ x(1 + v^{2}) = C_{1} $$
Back Substitution
Since $v = \dfrac{y}{x}$, we have
$$ x\left(1 + \frac{y^{2}}{x^{2}}\right) = C_{1} $$
Simplifying,
$$ \frac{x^{2} + y^{2}}{x} = C_{1} $$
or equivalently,
$$ \boxed{x^{2} + y^{2} = Cx} $$
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NCERT Question 5 : Solve the differential equation
$$x^{2}\frac{dy}{dx}=x^{2}-2y^{2}+xy.$$
Solution :
Check homogeneity and substitute
Rewrite in slope form:
$$\frac{dy}{dx}=\frac{x^{2}-2y^{2}+xy}{x^{2}}=1+ \frac{y}{x}-2\left(\frac{y}{x}\right)^{2}.$$
The right-hand side is a function of $v=\dfrac{y}{x}$ only, so the equation is homogeneous. Put
$$y=vx\qquad\Rightarrow\qquad \frac{dy}{dx}=v+x\frac{dv}{dx}.$$
Substitute into the DE:
$$
v+x\frac{dv}{dx}=1+v-2v^{2}.
$$
Hence
$$
x\frac{dv}{dx}=1-2v^{2}.
$$
Separate variables and integrate
Separate variables:
$$
\frac{dv}{1-2v^{2}}=\frac{dx}{x}.
$$
Integrate both sides. Using
$$
\int\frac{dv}{1-2v^{2}}=\frac{1}{2\sqrt{2}}\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|+C
$$
we get
$$
\frac{1}{2\sqrt{2}}\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=\ln|x|+C.
$$
Multiply by $2\sqrt{2}$ and exponentiate:
$$
\ln\left|\frac{1+\sqrt{2}v}{1-\sqrt{2}v}\right|=2\sqrt{2}\ln|x|+C_1
$$
$$
\frac{1+\sqrt{2}v}{1-\sqrt{2}v}=Cx^{2\sqrt{2}}\qquad(C=e^{C_1}>0).
$$
Back-substitute $v=\dfrac{y}{x}$
The implicit general solution is
$$
\boxed{\frac{1+\sqrt{2}\dfrac{y}{x}}{1-\sqrt{2}\dfrac{y}{x}}=Cx^{2\sqrt{2}}},
$$
or equivalently
$$
\boxed{\frac{x+\sqrt{2}y}{x-\sqrt{2}y}=Cx^{2\sqrt{2}}}.
$$
You can solve explicitly for $y$ if desired:
$$
\displaystyle y=x\cdot\frac{C x^{2\sqrt{2}}-1}{\sqrt{2}\big(1+C x^{2\sqrt{2}}\big)}.
$$
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NCERT Question 6 : Solve the differential equation
$$xdy – ydx = \sqrt{x^{2}+y^{2}}dx$$
Solution :
Given :
$$xdy – ydx = \sqrt{x^{2}+y^{2}}dx$$
(using homogeneous substitution)
Rearrange:
$$xdy = \bigl(y+\sqrt{x^{2}+y^{2}}\bigr)dx$$
so
$$\frac{dy}{dx}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}.$$
Let $y=vx$ (so $v=\dfrac{y}{x}$ and $dy/dx=v+xdv/dx$).
Also $\sqrt{x^{2}+y^{2}}=x\sqrt{1+v^{2}}$. Substituting gives
$$v+x\frac{dv}{dx}=v+\sqrt{1+v^{2}}.$$
Cancel $v$ and separate variables:
$$x\frac{dv}{dx}=\sqrt{1+v^{2}} \quad\Longrightarrow\quad \frac{dv}{\sqrt{1+v^{2}}}=\frac{dx}{x}.$$
Integrate both sides. Using $\displaystyle\int\frac{dv}{\sqrt{1+v^{2}}}=\ln!\big(v+\sqrt{1+v^{2}}\big)$,
$$\ln!\big(v+\sqrt{1+v^{2}}\big)=\ln|x|+C.$$
Exponentiate and rename the constant:
$$v+\sqrt{1+v^{2}}=Kx.$$
Substitute $v=\dfrac{y}{x}$ and multiply by $x$:
$$\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}=Kx \quad\Longrightarrow\quad y+\sqrt{x^{2}+y^{2}}=Kx^{2}.$$
Final Solution
$$\boxed{y+\sqrt{x^{2}+y^{2}}=C x^{2}}, \quad \text{where } C \text{ is an arbitrary constant.}$$
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