Answer to Question #132255 in Differential Equations for Rouhish Ray

Question

Solve: dy/dx + y = y^3(cosx - sinx)

Answer 1

Bernoulli differential equation

$$\dfrac{dy}{dx}+y=(\cos x-\sin x)y^3, n=3$$

Divide both sides by $y^3$

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

To find the solution, change the dependent variable from y to z, where $z=y^{1-3}=y^{-2}$

$$\dfrac{dz}{dx}=(-\dfrac{2}{y^3})\dfrac{dy}{dx}$$

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

$$\dfrac{dz}{dx}-2z=-2\cos x+2\sin x$$

$P(x)=-2$

We have the integrating factor

$$I(x)=e^{\int(-2)dx}=e^{-2x}$$

Then multiplying through by $I(x)$, we get

$$e^{-2x}\dfrac{dz}{dx}-2e^{-2x}z=2e^{-2x}(\sin x-\cos x)$$

$$\dfrac{d}{dx}(e^{-2x}z)=2e^{-2x}(\sin x-\cos x)$$

$$e^{-2x}z=2\int e^{-2x}(\sin x-\cos x)dx$$

$$\int e^{-2x}\sin xdx=-\cos x e^{-2x}-2\int e^{-2x}\cos xdx$$

$$\int e^{-2x}\cos xdx=\sin x e^{-2x}+2\int e^{-2x}\sin xdx$$

$$\int e^{-2x}\sin xdx=-\cos x e^{-2x}-2\sin x e^{-2x}-4\int e^{-2x}\sin xdx$$

$$5\int e^{-2x}\sin xdx=-\cos x e^{-2x}-2\sin x e^{-2x}$$

$$\int e^{-2x}\sin xdx=-\dfrac{1}{5}\cos x e^{-2x}-\dfrac{2}{5}\sin x e^{-2x}$$

$$\int e^{-2x}\cos xdx=\sin x e^{-2x}-\dfrac{2}{5}\cos x e^{-2x}-\dfrac{4}{5}\sin x e^{-2x}=$$

$$=-\dfrac{2}{5}\cos x e^{-2x}+\dfrac{1}{5}\sin x e^{-2x}$$

$$e^{-2x}z=-\dfrac{2}{5}\cos x e^{-2x}-\dfrac{4}{5}\sin x e^{-2x}+$$

$$+\dfrac{4}{5}\cos x e^{-2x}-\dfrac{2}{5}\sin x e^{-2x}+C=$$

$$=\dfrac{2}{5}\cos x e^{-2x}-\dfrac{6}{5}\sin x e^{-2x}+C$$

$$z=\dfrac{2}{5}\cos x -\dfrac{6}{5}\sin x +Ce^{2x}$$

$$\dfrac{1}{y^2}=\dfrac{2}{5}\cos x -\dfrac{6}{5}\sin x +Ce^{2x}$$

$$y^2=\dfrac{1}{\dfrac{2}{5}\cos x -\dfrac{6}{5}\sin x +Ce^{2x}}$$