Differential Equations Ques 57
- The curve amongst the family of curves represented by the differential equation, $\left(x^{2}-y^{2}\right) d x+2 x y d y=0$, which passes through $(1,1)$, is
(2019 Main, 10 Jan II)
(a) a circle with centre on the $Y$-axis
(b) a circle with centre on the $X$-axis
(c) an ellipse with major axis along the $Y$-axis
(d) a hyperbola with transverse axis along the $X$-axis.
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Answer:
Correct Answer: 57.(b)
Solution:
- Given differential equation is
$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$, which can be written as
$$ \frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y} $$
Put $y=v x \quad[\because$ it is in homogeneous form $]$
$\Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x}$
Now, differential equation becomes
$$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{v^{2} x^{2}-x^{2}}{2 x(v x)} \Rightarrow v+x \frac{d v}{d x}=\frac{\left(v^{2}-1\right) x^{2}}{2 v x^{2}} \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v=\frac{v^{2}-1-2 v^{2}}{2 v} \\ & \Rightarrow \quad x \frac{d v}{d x}=-\frac{1+v^{2}}{2 v} \Rightarrow \int \frac{2 v d v}{1+v^{2}}=-\int \frac{d x}{x} \end{aligned} $$
$\Rightarrow \quad \ln \left(1+v^{2}\right)=-\ln x-\ln C$
$$ \begin{array}{rrr} & & \because \int \frac{f^{\prime}(x)}{f(x)} d x \Rightarrow \ln |f(x)|+C \\ \Rightarrow \quad \ln \left|\left(1+v^{2}\right) C x\right|=0 & {[\because \ln A+\ln B=\ln A B]} \\ \Rightarrow \quad\left(1+v^{2}\right) C x=1 & {\left[\log _e x=0 \Rightarrow x=e^{0}=1\right]} \end{array} $$
Now, putting $v=\frac{y}{x}$, we get
$$ 1+\frac{y^{2}}{x^{2}} C x=1 \quad \Rightarrow C\left(x^{2}+y^{2}\right)=x $$
$\because$ The curve passes through $(1,1)$, so
$$ C(1+1)=1 \Rightarrow C=\frac{1}{2} $$
Thus, required curve is $x^{2}+y^{2}-2 x=0$, which represent a circle having centre $(1,0)$
$\therefore$ The solution of given differential equation represents a circle with centre on the $X$-axis.
BETA
