Solve e ^ y + ∫ (from 0 to y) of e ^t^2dt-x-1=0

$$e ^ { y } + \int _ { 0 } ^ { y } e ^ { t ^ { 2 } } d t - x - 1 = 0$$

Solve for x

$$x=\int _{0}^{y}e^{t^{2}}\mathrm{d}t+e^{y}-1$$

Steps for Solving Linear Equation

Subtract e^{y} from both sides. Anything subtracted from zero gives its negation.

$$\int _{0}^{y}e^{t^{2}}\mathrm{d}t-x-1=-e^{y}$$

Subtract \int _{0}^{y}e^{t^{2}}\mathrm{d}t from both sides.

$$-x-1=-e^{y}-\int _{0}^{y}e^{t^{2}}\mathrm{d}t$$

Add 1 to both sides.

$$-x=-e^{y}-\int _{0}^{y}e^{t^{2}}\mathrm{d}t+1$$

Reorder the terms.

$$-x=-\int _{0}^{y}e^{t^{2}}\mathrm{d}t-e^{y}+1$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{-\int _{0}^{y}e^{t^{2}}\mathrm{d}t-e^{y}+1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{-\int _{0}^{y}e^{t^{2}}\mathrm{d}t-e^{y}+1}{-1}$$

Divide -\int _{0}^{y}e^{t^{2}}\mathrm{d}t-e^{y}+1 by -1.

$$x=-\left(-\int _{0}^{y}e^{t^{2}}\mathrm{d}t-e^{y}+1\right)$$