If the solution of $\frac{d y}{d x}=\frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1 / y^2}}, y(0)=1$ is…
If the solution of $\frac{d y}{d x}=\frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1 / y^2}}, y(0)=1$ is $\frac{1}{y^2}=\log _e(f(x))$, then $f(x)=$
- $4+4 \sin \sqrt{x}$
- $e \sin \sqrt{x}$
- $1-4 \sin \sqrt{x}$
- $e-4 \sin \sqrt{x}$
Solution
$\frac{d y}{d x}=\frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1 / y^2}}$
$\Rightarrow \int \frac{e^{1 / y^2}}{y^3} d y=\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$
On putting $\frac{1}{y^2}=t \Rightarrow-\frac{2}{y^3} d y=d t$
and on putting $u=\sqrt{x} \Rightarrow d u=\frac{d x}{2 \sqrt{x}}$
we get $\int \frac{e^t d t}{(-2)}=\int 2 \cos u d u$
$\begin{aligned} & \Rightarrow \quad-\frac{1}{2}\left(e^t\right)=2 \sin u+k \\ & \Rightarrow \quad-\frac{1}{2}\left(e^{1 / y^2}\right)=2 \sin \sqrt{x}+k\end{aligned}$
$\because y(0)=1$ means at $x=0, y=1$
$\Rightarrow \quad-\frac{1}{2}\left(e^{1 / y^1}\right)=2 \sin \sqrt{0}+k$
$\Rightarrow \quad k=-\frac{e}{2}$
$\therefore \quad-\frac{1}{2}\left(e^{1 / y^2}\right)=2 \sin \sqrt{x}-\frac{e}{2}$
Takin, $\log$ both sides with base $e$,
$\log \left(\frac{e^{1 / y^2}}{2}\right)=\log \left[(2 \sin \sqrt{x})-\frac{e}{2}\right]$
$\begin{aligned} & \Rightarrow \quad \log \left(-\frac{1}{2}\right)+\log \left(e^{1 / y^2}\right)=\log \left(2 \sin \sqrt{x}-\frac{e}{2}\right) \\ & \Rightarrow \quad \log 1-\log (-2)+\frac{1}{y^2}=\log \left(2 \sin \sqrt{x}-\frac{e}{2}\right) \\ & \Rightarrow \quad \frac{1}{y^2}=\log \left[\left(2 \sin \sqrt{x}-\frac{e}{2}\right) \times(-2)\right] \\ & \Rightarrow \quad \frac{1}{y^2}=\log (e-4 \sin \sqrt{x})\end{aligned}$
$\therefore \quad f(x)=e-4 \sin \sqrt{x}$
Asked in: AP EAMCET 2022 (08 Jul Shift 2)
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