Differential Equations Ques 37
- Consider the differential equation, $y^{2} d x+x-\frac{1}{y} d y=0$. If value of $y$ is 1 when $x=1$, then the value of $x$ for which $y=2$, is
(2019 Main, 12 April I)
(a) $\frac{5}{2}+\frac{1}{\sqrt{e}}$
(b) $\frac{3}{2}-\frac{1}{\sqrt{e}}$
(c) $\frac{1}{2}+\frac{1}{\sqrt{e}}$
(d) $\frac{3}{2}-\sqrt{e}$
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Answer:
Correct Answer: 37.(b)
Solution:
Formula:
ELEMENTARY TYPES OF FIRST ORDER \& FIRST DEGREE DIFFERENTIAL EQUATIONS :
- Given differential equation is
$$ y^{2} d x+x-\frac{1}{y} \quad d y=0 $$
$\Rightarrow \frac{d x}{d y}+\frac{1}{y^{2}} x=\frac{1}{y^{3}}$, which is the linear differential equation of the form $\frac{d x}{d y}+P x=Q$.
Here, $P=\frac{1}{y^{2}}$ and $Q=\frac{1}{y^{3}}$
Now, IF $=e^{\int \frac{1}{y^{2}} d y}=e^{-\frac{1}{y}}$
$\therefore$ The solution of linear differential equation is
$$ \begin{aligned} x \cdot(IF) & \int(IF) d y+C \\ \Rightarrow x e^{-1 / y} & =\int \frac{1}{y^{3}} e^{-1 / y} d y+C \\ \therefore x e^{-1 / y} & =\int(-t) e^{t} d t+C \quad\left[\because \text { let }-\frac{1}{y}=t \Rightarrow+\frac{1}{y^{2}} d y=d t\right] \\ & =-t e^{t}+\int e^{t} d t+C \quad \text { [integration by parts] } \\ & =-t e^{t}+e^{t}+C \end{aligned} $$
$$ \Rightarrow x e^{-1 / y}=\frac{1}{y} e^{-1 / y}+e^{-1 / y}+C $$
Now, at $y=1$, the value of $x=1$, so
$$ 1 \cdot e^{-1}=e^{-1}+e^{-1}+C \Rightarrow C=-\frac{1}{e} $$
On putting the value of $C$, in Eq. (i), we get
$$ x=\frac{1}{y}+1-\frac{e^{1 / y}}{e} $$
So, at $y=2$, the value of $x=\frac{1}{2}+1-\frac{e^{1 / 2}}{e}=\frac{3}{2}-\frac{1}{\sqrt{e}}$
BETA
