Answer to Question #103557 in Mechanics | Relativity for Praveen

Question

Solve the following ordinary differential equations

(a) dy/dx+ycotx=power of e is cosx
For x=π/2 y=-2

(b) d square y/dx square + dy/dx + y =0

Answer 1

a)

$$\frac{dy}{dx}+y\cot x=e^{cosx}$$

$$p=\cot x, q=e^{cosx}$$

$$IF=\exp\int pdx=\exp\int \cot xdx=\exp (\ln\sin x)$$

$$IF=\sin x$$

$$y\sin x=\int e^{cosx}\sin x dx=c-e^{cosx}$$

$$-2\sin 90=c-e^{cos90}\to c=-1$$

$$y=-\frac{1+e^{cos90}}{\sin x}$$

b)

$$\frac{d^2y}{dx^2}+\frac{dy}{dx}+y=0\to y=e^{\lambda x}$$

$$\lambda ^2+\lambda +1=0$$

$$\lambda_{1,2}=\frac{-1\pm \sqrt{3}}{2}$$

$$y=c_1e^{-\frac{x}{2}}\sin\left (\frac{\sqrt{3}}{2}x\right)+c_2e^{-\frac{x}{2}}\cos\left (\frac{\sqrt{3}}{2}x\right)$$