# \frac{dy}{dx}= (xy)^2

## Question:

Solve {eq}\frac{dy}{dx}= (xy)^2{/eq}

## Solve a Differential Equation:

To solve a differential equation, {eq}\dfrac{dy}{dx} = f(x,y) {/eq}, where the variable {eq}x {/eq} and {eq}y {/eq} in {eq}f(x,y) {/eq} can be separated, such that {eq}f(x,y)=g(x)h(y) {/eq}, we group terms of {eq}x {/eq} and {eq}y {/eq} in the different side and then evaluate it by applying integral, {eq}\int \dfrac{dy}{h(y)} = \int g(x) dx {/eq}.

To solve {eq}\dfrac{dy}{dx}= (xy)^2{/eq}, we separate the factor of variable {eq}x {/eq} and variable {eq}y {/eq}.

{eq}\displaystyle \frac{dy}{dx}= (xy)^2 \\ \Rightarrow \displaystyle \frac{dy}{dx}= x^2y^2 \\ \Rightarrow \displaystyle \frac{1}{y^2} \, dy = x^2 \, dx \\ {/eq}

Then, we take integral of both sides,

{eq}\displaystyle \int \frac{1}{y^2} \, dy = \int x^2 \, dx \\ \Rightarrow \displaystyle \int y^{-2} \, dy = \int x^2 \, dx \\ \Rightarrow \displaystyle -1(y^{-1}) = \frac{1}{3}x^3 \quad \text{[Apply integration formula, } \int x^n \, dx = \frac{1}{n+1}x^{n+1} \text{]} \\ \Rightarrow \displaystyle - \frac{1}{y} = \frac{1}{3}x^3 \\ \Rightarrow \displaystyle -y = \frac{3}{x^3} \\ \Rightarrow \displaystyle \color{blue}{y = -\frac{3}{x^3} + C} \quad \text{[Add a constant C to the final result of indefinite integral]}\\ {/eq}