Solve ∫ sqrt{x}+x^2+x^2=θy

$$\int \sqrt { x } + x ^ { 2 } + x ^ { 2 } = \theta y$$

Solve for y

$$\left\{\begin{matrix}y=\frac{2x^{3}+2x^{\frac{3}{2}}}{3\theta }+\frac{С}{\theta }\text{, }&\theta \neq 0\text{ and }x\geq 0\\y\in \mathrm{R}\text{, }&С=\frac{-2x^{3}-2x^{\frac{3}{2}}}{3}\text{ and }x\geq 0\text{ and }\theta =0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine x^{2} and x^{2} to get 2x^{2}.

$$\int \sqrt{x}+2x^{2}\mathrm{d}x=\theta y$$

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

$$\theta y=\int \sqrt{x}+2x^{2}\mathrm{d}x$$

The equation is in standard form.

$$\theta y=\frac{2x^{3}}{3}+x^{\frac{3}{2}}+С$$

Divide both sides by \theta .

$$\frac{\theta y}{\theta }=\frac{\frac{2x^{3}}{3}+x^{\frac{3}{2}}+С}{\theta }$$

Dividing by \theta undoes the multiplication by \theta .

$$y=\frac{\frac{2x^{3}}{3}+x^{\frac{3}{2}}+С}{\theta }$$

Divide x^{\frac{3}{2}}+\frac{2x^{3}}{3}+С by \theta .

$$y=\frac{2x^{3}+3x^{\frac{3}{2}}+3С}{3\theta }$$

Solve for θ

$$\left\{\begin{matrix}\theta =\frac{2x^{3}+2x^{\frac{3}{2}}}{3y}+\frac{С}{y}\text{, }&y\neq 0\text{ and }x\geq 0\\\theta \in \mathrm{R}\text{, }&С=\frac{-2x^{3}-2x^{\frac{3}{2}}}{3}\text{ and }x\geq 0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine x^{2} and x^{2} to get 2x^{2}.

$$\int \sqrt{x}+2x^{2}\mathrm{d}x=\theta y$$

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

$$\theta y=\int \sqrt{x}+2x^{2}\mathrm{d}x$$

The equation is in standard form.

$$y\theta =\frac{2x^{3}}{3}+x^{\frac{3}{2}}+С$$

Divide both sides by y.

$$\frac{y\theta }{y}=\frac{\frac{2x^{3}}{3}+x^{\frac{3}{2}}+С}{y}$$

Dividing by y undoes the multiplication by y.

$$\theta =\frac{\frac{2x^{3}}{3}+x^{\frac{3}{2}}+С}{y}$$

Divide x^{\frac{3}{2}}+\frac{2x^{3}}{3}+С by y.

$$\theta =\frac{2x^{3}+3x^{\frac{3}{2}}+3С}{3y}$$