Solve (1+t^2){g(t+1)-g(t)}=2t

$$( 1 + t ^ { 2 } ) \{ g ( t + 1 ) - g ( t ) \} = 2 t$$

Solve for g (complex solution)

$$g=\frac{2t}{t^{2}+1}$$

$$t\neq -i\text{ and }t\neq i$$

Steps for Solving Linear Equation

Use the distributive property to multiply g by t+1.

$$\left(1+t^{2}\right)\left(gt+g-gt\right)=2t$$

Combine gt and -gt to get 0.

$$\left(1+t^{2}\right)g=2t$$

The equation is in standard form.

$$\left(t^{2}+1\right)g=2t$$

Divide both sides by 1+t^{2}.

$$\frac{\left(t^{2}+1\right)g}{t^{2}+1}=\frac{2t}{t^{2}+1}$$

Dividing by 1+t^{2} undoes the multiplication by 1+t^{2}.

$$g=\frac{2t}{t^{2}+1}$$

Solve for g

$$g=\frac{2t}{t^{2}+1}$$

Steps for Solving Linear Equation

Use the distributive property to multiply g by t+1.

$$\left(1+t^{2}\right)\left(gt+g-gt\right)=2t$$

Combine gt and -gt to get 0.

$$\left(1+t^{2}\right)g=2t$$

The equation is in standard form.

$$\left(t^{2}+1\right)g=2t$$

Divide both sides by 1+t^{2}.

$$\frac{\left(t^{2}+1\right)g}{t^{2}+1}=\frac{2t}{t^{2}+1}$$

Dividing by 1+t^{2} undoes the multiplication by 1+t^{2}.

$$g=\frac{2t}{t^{2}+1}$$

Solve for t (complex solution)

$$\left\{\begin{matrix}t=\frac{\sqrt{1-g^{2}}+1}{g}\text{; }t=\frac{-\sqrt{1-g^{2}}+1}{g}\text{, }&g\neq 0\\t=0\text{, }&g=0\end{matrix}\right.$$

Solve for t

$$\left\{\begin{matrix}t=\frac{\sqrt{1-g^{2}}+1}{g}\text{; }t=\frac{-\sqrt{1-g^{2}}+1}{g}\text{, }&g\neq 0\text{ and }|g|\leq 1\\t=0\text{, }&g=0\end{matrix}\right.$$