Hi Guys
i need help to solve this differentail equation :
cos^2 x sinx d' =cos^3x.y+1
i tried many times to solve this question but i could not
so please help me to solve question
You probably haven't received any help yet because the way you have posted your DE is difficult to read.
Is this your DE?
$\displaystyle \begin{align*} \cos^2{(x)} \sin{(x)} \,\frac{\mathrm{d}y}{\mathrm{d}x} = y\cos^3{(x)} + 1 \end{align*}$
Rewrite it as
$\displaystyle \begin{align*} \cos^2{(x)}\sin{(x)}\,\frac{\mathrm{d}y}{\mathrm{d }x} &= y\,\cos^3{(x)} + 1 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y\,\cos^3{(x)} + 1}{\cos^2{(x)}\sin{(x)}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= y\,\frac{\cos{(x)}}{\sin{(x)}} + \frac{1}{\cos^2{(x)}\sin{(x)}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{\cos{(x)}}{\sin{(x)}}\,y &= \frac{1}{\cos^2{(x)}\sin{(x)}} \end{align*}$
This is now first order linear, so you can solve the DE using an integrating factor.