Solve a_{1}(x)y^prime+a_{0}(x)y=g(x)

$$a _ { 1 } ( x ) y ^ { \prime } + a _ { 0 } ( x ) y = g ( x )$$

Solve for a_0 (complex solution)

$$\left\{\begin{matrix}a_{0}=\frac{g}{y}\text{, }&y\neq 0\\a_{0}\in \mathrm{C}\text{, }&x=0\text{ or }\left(g=0\text{ and }y=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y) from both sides.

$$a_{0}xy=gx-a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)$$

Reorder the terms.

$$a_{0}xy=-a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)+gx$$

The equation is in standard form.

$$xya_{0}=gx$$

Divide both sides by xy.

$$\frac{xya_{0}}{xy}=\frac{gx}{xy}$$

Dividing by xy undoes the multiplication by xy.

$$a_{0}=\frac{gx}{xy}$$

Divide gx by xy.

$$a_{0}=\frac{g}{y}$$

Solve for a_1 (complex solution)

$$a_{1}\in \mathrm{C}$$

$$g=a_{0}y\text{ or }x=0$$

Steps for Solving Linear Equation

Subtract a_{0}xy from both sides.

$$a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)=gx-a_{0}xy$$

Reorder the terms.

$$a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)=-a_{0}xy+gx$$

The equation is in standard form.

$$0=gx-a_{0}xy$$

This is false for any a_{1}.

$$a_{1}\in$$

Solve for a_0

$$\left\{\begin{matrix}a_{0}=\frac{g}{y}\text{, }&y\neq 0\\a_{0}\in \mathrm{R}\text{, }&x=0\text{ or }\left(g=0\text{ and }y=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y) from both sides.

$$a_{0}xy=gx-a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)$$

Reorder the terms.

$$a_{0}xy=-a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)+gx$$

The equation is in standard form.

$$xya_{0}=gx$$

Divide both sides by xy.

$$\frac{xya_{0}}{xy}=\frac{gx}{xy}$$

Dividing by xy undoes the multiplication by xy.

$$a_{0}=\frac{gx}{xy}$$

Divide gx by xy.

$$a_{0}=\frac{g}{y}$$

Solve for a_1

$$a_{1}\in \mathrm{R}$$

$$g=a_{0}y\text{ or }x=0$$

Steps for Solving Linear Equation

Subtract a_{0}xy from both sides.

$$a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)=gx-a_{0}xy$$

Reorder the terms.

$$a_{1}x\frac{\mathrm{d}}{\mathrm{d}x}(y)=-a_{0}xy+gx$$

The equation is in standard form.

$$0=gx-a_{0}xy$$

This is false for any a_{1}.

$$a_{1}\in$$