Solve y=e^x(Acosx+bsinx

$$y = e ^ { x } ( A \cos x + b \sin x$$

Solve for A

$$\left\{\begin{matrix}A=\frac{-b\sin(x)e^{x}+y}{\cos(x)e^{x}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}+\frac{\pi }{2}\\A\in \mathrm{R}\text{, }&y=b\sin(x)e^{x}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply e^{x} by A\cos(x)+b\sin(x).

$$y=e^{x}A\cos(x)+e^{x}b\sin(x)$$

Swap sides so that all variable terms are on the left hand side.

$$e^{x}A\cos(x)+e^{x}b\sin(x)=y$$

Subtract e^{x}b\sin(x) from both sides.

$$e^{x}A\cos(x)=y-e^{x}b\sin(x)$$

The equation is in standard form.

$$\cos(x)e^{x}A=-b\sin(x)e^{x}+y$$

Divide both sides by e^{x}\cos(x).

$$\frac{\cos(x)e^{x}A}{\cos(x)e^{x}}=\frac{-b\sin(x)e^{x}+y}{\cos(x)e^{x}}$$

Dividing by e^{x}\cos(x) undoes the multiplication by e^{x}\cos(x).

$$A=\frac{-b\sin(x)e^{x}+y}{\cos(x)e^{x}}$$

Divide y-e^{x}b\sin(x) by e^{x}\cos(x).

$$A=\frac{\frac{y}{e^{x}}-b\sin(x)}{\cos(x)}$$

Solve for b

$$\left\{\begin{matrix}b=\frac{-A\cos(x)e^{x}+y}{\sin(x)e^{x}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\\b\in \mathrm{R}\text{, }&y=A\cos(x)e^{x}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply e^{x} by A\cos(x)+b\sin(x).

$$y=e^{x}A\cos(x)+e^{x}b\sin(x)$$

Swap sides so that all variable terms are on the left hand side.

$$e^{x}A\cos(x)+e^{x}b\sin(x)=y$$

Subtract e^{x}A\cos(x) from both sides.

$$e^{x}b\sin(x)=y-e^{x}A\cos(x)$$

The equation is in standard form.

$$\sin(x)e^{x}b=-A\cos(x)e^{x}+y$$

Divide both sides by e^{x}\sin(x).

$$\frac{\sin(x)e^{x}b}{\sin(x)e^{x}}=\frac{-A\cos(x)e^{x}+y}{\sin(x)e^{x}}$$

Dividing by e^{x}\sin(x) undoes the multiplication by e^{x}\sin(x).

$$b=\frac{-A\cos(x)e^{x}+y}{\sin(x)e^{x}}$$

Divide y-e^{x}A\cos(x) by e^{x}\sin(x).

$$b=\frac{\frac{y}{e^{x}}-A\cos(x)}{\sin(x)}$$