Answer to Question #294062 in Differential Equations for iman munaim

Question

a)solve dy/dx - 3ye^3x =y. Given that y=1 when x=0

b)find the solution of sinx dy/dx + y cosx = sin^3 x cosx

Answer 1

a)

$$y'-3ye^{3x}=y$$

$$y'=y(1+3e^{3x})$$

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

Integrate

$$\int \dfrac{dy}{y}=\int (1+3e^{3x})dx$$

$$\ln (|y|)=x+e^{3x}+\ln C$$

$$y=Ce^{x+e^{3x}}$$

$1(0)=1$

$$1=Ce^{0+e^{3(0)}}=>C=e^{-1}$$

$$y=e^{x+e^{3x}-1}$$

b)

$$d(y\sin x)=\sin^3x\cos xdx$$

Integrate

$$\int d(y\sin x)=\int \sin^3x\cos xdx$$

$$y\sin x=\dfrac{\sin^4 x}{4}+C$$

$$y=\dfrac{\sin^3 x}{4}+\dfrac{C}{\sin x}$$