Practice: Solving separable differential equation

Calculus Level 1

What is the solution to the differential equation $\frac{dy}{dx}=\frac{x^2}{y^2}$ with initial condition $y(0)=2?$

y=\sqrt [4 ]{ x+4 }

y=\sqrt [3 ]{ 2x^2+2 }

y=\sqrt [3 ]{ x^3+8 }

y=\sqrt [2 ]{ x^2+10 }

1 solution

L N
Jun 6, 2014

This a separable differential equation. Hence, we are allowed to do: $y^2 dy = x^2 dx$ then we integrate both sides to get: $\frac{y^3}{3} = \frac{x^3}{3}$ $y^3 = x^3 + C$ $2^3 = 0^3 + C$ $C = 8$ Then we further simplify: $y = (x^3+8)^{\frac{1}{3}}$

Practice: Solving separable differential equation

1 solution

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