Question

If $\frac{dy}{dx}+\frac{y}{\sqrt{x^2+a^2}}=3x,y\left(0\right)=a^2$, then the value of $\frac{y\left(\sqrt{3}a\right)}{a^2}.\frac{8-3\sqrt{3}}{2-\sqrt{3}}$ is equal to

• A. 64
• B. 37
• C. 23
• D. 19

Answer 1

The correct option is B 37

$\frac{dy}{dx}+\frac{y}{\sqrt{x^2+a^2}}=3x$ ...(1)
Let $u=e^{\int \frac{1}{\sqrt{a^2+x^2}}dx}=\sqrt{a^2+x^2}+x$
Multiplying both sides of (1) by $u$
$(\sqrt{a^2+x^2}+x)\frac{dy}{dx}+(\sqrt{a^2+x^2}+x)\frac{y}{\sqrt{x^2+a^2}}=3x(\sqrt{a^2+x^2}+x)\Rightarrow (\sqrt{a^2+x^2}+x)\frac{dy}{dx}+y\frac{d}{dx}(\sqrt{a^2+x^2}+x)=3x(\sqrt{a^2+x^2}+x)$
Using $g\frac{df}{dx}+f\frac{dg}{dx}=\frac{d}{dx}\left(fg\right)$
$\frac{d}{dx}\left((\sqrt{a^2+x^2}+x)y\right)=3x(\sqrt{a^2+x^2}+x)$
Integrating both sides w.r.t.x
$(\sqrt{a^2+x^2}+x)y=x^3+{(a^2+x^2)}^{3/2}+c$
For $y\left(0\right)=a^2\Rightarrow (\sqrt{a^2+0^2}+0)a^2=0^3+{(a^2+0^2)}^{3/2}+c\Rightarrow a^3=a^3+c\Rightarrow c=0$
Then for $x=\sqrt{3}a$
$(\sqrt{a^2+{3a}^2}+\sqrt{3}a)y\left(\sqrt{3}a\right)={\left(\sqrt{3}a\right)}^3+{(a^2+{3a}^2)}^{3/2}\Rightarrow (2a+\sqrt{3}a)y\left(\sqrt{3}a\right)=3\sqrt{3}{a}^3+8a^3\Rightarrow y\left(\sqrt{3}a\right)=\frac{3\sqrt{3}{a}^3+8a^3}{2a+\sqrt{3}a}$
Therefore,
$\frac{y\left(\sqrt{3}a\right)}{a^2}\frac{8-3\sqrt{3}}{2-\sqrt{3}}=\frac{3\sqrt{3}{a}^3+8a^3}{2a+\sqrt{3}a}\times \frac{1}{a^2}\times \frac{8-3\sqrt{3}}{2-\sqrt{3}}=37$