Solve y=axe^xsin2x+bxe^xcos2x

$$y = a x e ^ { x } \sin 2 x + b x e ^ { x } \cos 2 x$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{bxe^{\left(1-2i\right)x}+bxe^{\left(1+2i\right)x}-2y}{x\left(ie^{\left(1-2i\right)x}-ie^{\left(1+2i\right)x}\right)}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}\\a\in \mathrm{C}\text{, }&y=\frac{bx\left(e^{\left(1-2i\right)x}+e^{\left(1+2i\right)x}\right)}{2}\text{ and }\left(\left(y=\frac{bx\left(e^{\left(1-2i\right)x}+e^{\left(1+2i\right)x}\right)}{2}\text{ and }ixe^{\left(1-2i\right)x}-ixe^{\left(1+2i\right)x}=0\right)\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$axe^{x}\sin(2x)+bxe^{x}\cos(2x)=y$$

Subtract bxe^{x}\cos(2x) from both sides.

$$axe^{x}\sin(2x)=y-bxe^{x}\cos(2x)$$

The equation is in standard form.

$$x\sin(2x)e^{x}a=-bx\cos(2x)e^{x}+y$$

Divide both sides by xe^{x}\sin(2x).

$$\frac{x\sin(2x)e^{x}a}{x\sin(2x)e^{x}}=\frac{-bx\cos(2x)e^{x}+y}{x\sin(2x)e^{x}}$$

Dividing by xe^{x}\sin(2x) undoes the multiplication by xe^{x}\sin(2x).

$$a=\frac{-bx\cos(2x)e^{x}+y}{x\sin(2x)e^{x}}$$

Divide y-bxe^{x}\cos(2x) by xe^{x}\sin(2x).

$$a=\frac{\frac{y}{xe^{x}}-b\cos(2x)}{\sin(2x)}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\frac{iaxe^{\left(1-2i\right)x}-iaxe^{\left(1+2i\right)x}-2y}{x\left(e^{\left(1-2i\right)x}+e^{\left(1+2i\right)x}\right)}\text{, }&x\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\text{ and }xe^{\left(1-2i\right)x}+xe^{\left(1+2i\right)x}\neq 0\\b\in \mathrm{C}\text{, }&\left(y=0\text{ and }x=0\right)\text{ or }\left(y=\frac{ax\left(ie^{\left(1-2i\right)x}-ie^{\left(1+2i\right)x}\right)}{2}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$axe^{x}\sin(2x)+bxe^{x}\cos(2x)=y$$

Subtract axe^{x}\sin(2x) from both sides.

$$bxe^{x}\cos(2x)=y-axe^{x}\sin(2x)$$

The equation is in standard form.

$$x\cos(2x)e^{x}b=-ax\sin(2x)e^{x}+y$$

Divide both sides by xe^{x}\cos(2x).

$$\frac{x\cos(2x)e^{x}b}{x\cos(2x)e^{x}}=\frac{-ax\sin(2x)e^{x}+y}{x\cos(2x)e^{x}}$$

Dividing by xe^{x}\cos(2x) undoes the multiplication by xe^{x}\cos(2x).

$$b=\frac{-ax\sin(2x)e^{x}+y}{x\cos(2x)e^{x}}$$

Divide y-axe^{x}\sin(2x) by xe^{x}\cos(2x).

$$b=\frac{\frac{y}{xe^{x}}-a\sin(2x)}{\cos(2x)}$$

Solve for a

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

Steps for Solving Linear Equation

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

$$axe^{x}\sin(2x)+bxe^{x}\cos(2x)=y$$

Subtract bxe^{x}\cos(2x) from both sides.

$$axe^{x}\sin(2x)=y-bxe^{x}\cos(2x)$$

The equation is in standard form.

$$x\sin(2x)e^{x}a=-bx\cos(2x)e^{x}+y$$

Divide both sides by xe^{x}\sin(2x).

$$\frac{x\sin(2x)e^{x}a}{x\sin(2x)e^{x}}=\frac{-bx\cos(2x)e^{x}+y}{x\sin(2x)e^{x}}$$

Dividing by xe^{x}\sin(2x) undoes the multiplication by xe^{x}\sin(2x).

$$a=\frac{-bx\cos(2x)e^{x}+y}{x\sin(2x)e^{x}}$$

Divide y-bxe^{x}\cos(2x) by xe^{x}\sin(2x).

$$a=\frac{-bx\cos(2x)+\frac{y}{e^{x}}}{x\sin(2x)}$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{ax\sin(2x)e^{x}-y}{x\cos(2x)e^{x}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi n_{1}}{2}+\frac{\pi }{4}\text{ and }x\neq 0\\b\in \mathrm{R}\text{, }&\left(y=0\text{ and }x=0\right)\text{ or }\left(y=ax\sin(2x)e^{x}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{4}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$axe^{x}\sin(2x)+bxe^{x}\cos(2x)=y$$

Subtract axe^{x}\sin(2x) from both sides.

$$bxe^{x}\cos(2x)=y-axe^{x}\sin(2x)$$

The equation is in standard form.

$$x\cos(2x)e^{x}b=-ax\sin(2x)e^{x}+y$$

Divide both sides by xe^{x}\cos(2x).

$$\frac{x\cos(2x)e^{x}b}{x\cos(2x)e^{x}}=\frac{-ax\sin(2x)e^{x}+y}{x\cos(2x)e^{x}}$$

Dividing by xe^{x}\cos(2x) undoes the multiplication by xe^{x}\cos(2x).

$$b=\frac{-ax\sin(2x)e^{x}+y}{x\cos(2x)e^{x}}$$

Divide y-axe^{x}\sin(2x) by xe^{x}\cos(2x).

$$b=\frac{-ax\sin(2x)+\frac{y}{e^{x}}}{x\cos(2x)}$$