Question

In Problems 1–22 Solve the given differential equation by separation of variables.

${\mathbf{\text{Sin(3x)dx+2yCos}}}^{\mathbf{\text{3}}}\mathbf{\text{(3x)dy=0}}$

Short answer

$y^2=-\frac{1}{6}{\mathrm{Sec}}^2\left(3x\right)$

Step-by-step solution

Step 1:Definition of separable equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Step 2:Separate variables

Given equation,

$\mathrm{Sin}\left(3x\right)dx+2y{\mathrm{Cos}}^3\left(3x\right)dy=0$

Move$\mathrm{Sin}\left(3x\right)dx$to the right side.

$2y{\mathrm{Cos}}^3\left(3x\right)dy=-\mathrm{Sin}\left(3x\right)dx$

Divide both sides by${\mathrm{Cos}}^3\left(3x\right)$

$2ydy=-\frac{\mathrm{Sin}\left(3x\right)}{{\mathrm{Cos}}^3\left(3x\right)}dx$

$\frac{\mathrm{Sin}\left(3x\right)}{\mathrm{Cos}\left(3x\right)}=tan\left(3x\right)\text{and}\frac{1}{{\mathrm{Cos}}^2\left(3x\right)}={\mathrm{Sec}}^2\left(3x\right)$

Step 3:Integration

$\int 2ydy=-\int tan\left(3x\right){\mathrm{Sec}}^2\left(3x\right)dx$

Trig identity$\frac{d}{dx}\mathrm{Sec}\left(x\right)=\mathrm{Sec}\left(x\right)tan\left(x\right)$

$u=\mathrm{Sec}\left(3x\right)\to du=3\mathrm{Sec}\left(3x\right)tan\left(3x\right)dx$

Integrate.

$\int 2ydy=-\frac{1}{3}\int udu$

Plug inand then you're done.

$y^2=-\frac{1}{6}{u}^2$

$y^2=-\frac{1}{6}{\mathrm{Sec}}^2\left(3x\right)$

Hence the solution is${\text{y}}^{\text{2}}\text{=-}\frac{\text{1}}{\text{6}}{\text{Sec}}^{\text{2}}\text{(3x)}$