Let y=y(x) be the solution of the differential equation [ x (d y)/(d x)- 2 y=x^3(2-x^3) ^2…

Question

Let $y=y(x)$ be the solution of the differential equation
$$ x \frac{d y}{d x}-\sin 2 y=x^3\left(2-x^3\right) \cos ^2 y, x \neq 0 . $$
If $y(2)=0$, then $\tan (y(1))$ is equal to

Accepted Answer

Given equation is :

$x\frac{dy}{dx}-\sin 2y={x}^{3}(2-{x}^{3}){\cos }^{2}y,\, x
e 0$

Divide by $x{\cos }^{2}y$

${\sec }^{2}y\frac{dy}{dx}-\frac{\sin 2y}{x{\cos }^{2}y}={x}^{2}(2-{x}^{3})$

${\sec }^{2}y\frac{dy}{dx}-\frac{2	an y}{x}=2{x}^{2}-{x}^{5}$

Substitute $	an y=t$ and differentiate w.r.t $x$

${\sec }^{2}y\frac{dy}{dx}=\frac{dt}{dx}$

Equation (1) becomes linear differential equation.

$\frac{dt}{dx}-\frac{2}{x}t=2{x}^{2}-{x}^{5}$

Integrating factor $IF={e}^{\int (-\frac{2}{x})dx}={e}^{-2\ln x}={e}^{\ln (\frac{1}{{x}^{2}})}=\frac{1}{{x}^{2}}$

Solution of linear differential equation :

$t\cdot IF=\int (IF)(2{x}^{2}-{x}^{5})dx$

$\Rightarrow$ $t\cdot \frac{1}{{x}^{2}}=\int \frac{1}{{x}^{2}}(2{x}^{2}-{x}^{5})dx$

$\Rightarrow$ $\frac{t}{{x}^{2}}=\int (2-{x}^{3})dx=2x-\frac{{x}^{4}}{4}+C$

put value of $t=	an y$

$\frac{	an y}{{x}^{2}}=2x-\frac{{x}^{4}}{4}+C$

using condition $y(2)=0$ to find $C$.

$\frac{	an (0)}{{2}^{2}}=2	imes 2-\frac{{2}^{4}}{4}+C\implies C=0$

So, solution of differential equation is

$	an y=2{x}^{3}-\frac{{x}^{6}}{4}$

to calculate $	an (y(1))$, put $x=1$ in the solution.

$	an (y(1))=2(1{)}^{3}-\frac{(1{)}^{6}}{4}$

$=2-\frac{1}{4}=\frac{7}{4}$

Solutions · 1