Answer to Question #163428 in Calculus for Sai preeth varma Lolakpuri

Question

solve the differential equation dy/dx + y cos x = y^3sin2x

Answer 1

$\frac{dy}{dx}+y\cos{x}=y^3\sin{2x}$

Solution:

Divide the left and right sides of the equation by $y^3$ :

$\frac{1}{y^3}\frac{dy}{dx}+\frac{1}{y^2}\cos{x}=\sin{2x}$

$\frac{1}{y^2}=z$ , $z'=-2\frac{y'}{y^3}$ , $\frac{y'}{y^3}=-\frac{z'}{2}$ .

$-\frac{z'}{2}+z\cos{x}=\sin{2x}$

$z'-2z\cos{x}=-2\sin{2x}$

$z=uv$ , $z'=u'v+v'u$

$u'v+v'u-2uv\cos{x}=-2\sin{2x}$

$u'v+u(v'-2v\cos{x})=-2\sin{2x}$

Let's compose and solve the system:

$\begin{cases} v'-2v\cos{x}=0 \\ u'v=-2\sin{2x} \end{cases}$

From the first equation:

$\frac{dv}{v}=2\cos{x}dx$

$\ln|v|=2\sin{x}$

$v=e^{2\sin{x}}$

Substitute $v$ into the second equation:

$u'e^{2\sin{x}}=-2\sin{2x}$

$du=-\frac{2\sin{2x}dx}{e^{2\sin{x}}}$

$\displaystyle u=-\int\frac{4\sin{x}\cos{x}dx}{e^{2\sin{x}}}=-\int\frac{4\sin{x}d\sin{x}}{e^{2\sin{x}}}=$

$\displaystyle\frac{2\sin{x}+1}{e^{2\sin{x}}}+C$

$z=uv=e^{2\sin{x}}(\displaystyle\frac{2\sin{x}+1}{e^{2\sin{x}}}+C)=$

$\displaystyle2\sin{x}+1+Ce^{2\sin{x}}$

$\frac{1}{y^2}=\displaystyle2\sin{x}+1+Ce^{2\sin{x}}$

Answer: $\frac{1}{y^2}=\displaystyle2\sin{x}+1+Ce^{2\sin{x}}$ .