Differential Equations Question 7
Question 7 - 30 January - Shift 1
Let the solution curve $y=y(x)$ of the differential
equation
$ \frac{d y}{d x}-\frac{3 x^{5} \tan ^{-1}(x^{3})}{(1+x^{6})^{\frac{3}{2}}} y=2 x $ also it passes through the origin. Then $y(1)$ is equal to :
(1) $\exp (\frac{4-\pi}{4 \sqrt{2}})$
(2) $\exp (\frac{\pi-4}{4 \sqrt{2}})$
(3) $\exp (\frac{1-\pi}{4 \sqrt{2}})$
(4) $\exp (\frac{4+\pi}{4 \sqrt{2}})$
Show Answer
Answer: (1)
Solution:
Formula: Linear differential equations of first order
$\frac{d y}{d x}+(\frac{-3 x^{5} \tan ^{-1} x^{3}}{(1+x^{6})^{3 / 2}}) y=2 e^{{\frac{x-\tan x}{\sqrt{1+x^{6}}}}}$
I.F. $=e^{\int \frac{-3 x^{5} \tan ^{-1} x^{3}}{(1+x^{6})^{3 / 2}} d x}$
$=e^{\frac{\tan ^{-1} x^{3}-x^{3}}{\sqrt{1+x^{6}}}}$
Solution of differential equation
$y \cdot e^{\frac{\tan ^{-1} x^{3}-x^{3}}{\sqrt{1+x^{6}}}}=\int 2 x e^{(\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{1+x^{6}}})} \cdot e^{(\frac{\tan ^{-1}(x^{3})-x^{3}}{\sqrt{1+x^{6}}})} d x$
$=\int 2 xdx+c$
$y \cdot e^{\frac{\tan ^{-1} x^{3}-x^{3}}{\sqrt{1+x^{6}}}}=x^{2}+c$
Also it passes through origin
$c=0$
$y(1) \cdot e^{\frac{\tan ^{-1}(1)-1}{\sqrt{2}}}=1$
$y(1) \cdot e^{\frac{\frac{\pi}{4}-1}{\sqrt{2}}}=1$
$y(1) \cdot e^{\frac{\pi-4}{4 \sqrt{2}}}=1$
$y(1)=\frac{1}{e^{\frac{\pi-4}{4 \sqrt{2}}}}=e^{\frac{4-\pi}{4 \sqrt{2}}}$
