Question

Find the general solution of the given differential equation. cos^2(x)sin(x)dy/dx+(cos^3(x))y=1 g

Answer 1

If the given differential equation is

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

then multiply both sides by $\frac1{\cos^2(x)}$ :

$\sin(x) \dfrac{dy}{dx} + \cos(x) y = \sec^2(x)$

The left side is the derivative of a product,

$\dfrac{d}{dx}\left[\sin(x)y\right] = \sec^2(x)$

Integrate both sides with respect to $x$, recalling that $\frac{d}{dx}\tan(x) = \sec^2(x)$ :

$\displaystyle \int \frac{d}{dx}\left[\sin(x)y\right] \, dx = \int \sec^2(x) \, dx$

$\sin(x) y = \tan(x) + C$

Solve for $y$ :

$\boxed{y = \sec(x) + C \csc(x)}

which follows from [tex]\tan(x)=\frac{\sin(x)}{\cos(x)}$.