Chalk−1

Math · Calculus II · Worked example

Solve a separable differential equation

Solve dy/dx = 2xy with the initial condition y(0) = 3.

Separate the variables

For y ≠ 0, divide by y and multiply by dx: dy/y = 2x dx.

Integrate both sides

∫dy/y = ln|y| and ∫2x dx = x², so ln|y| = x² + C, with one constant.

Solve for y

Exponentiate: |y| = e^C·e^(x²), so y = Ae^(x²) with A = ±e^C. The equilibrium y = 0, lost when dividing by y, is the case A = 0, so y = Ae^(x²) covers every solution.

Apply the initial condition

y(0) = Ae⁰ = A, so A = 3.

y⁢(x)=3ex2y⁢(0)=3

Check by differentiating

y′ = 6xe^(x²), which is 2x times y.

2⁢x⋅3ex26⁢xex2

Result

y = 3e^(x²).

Your turn

Solve dy/dx = 3x²y with y(0) = 2.

Show the answer and explanation

y = 2e^(x³).

dy/y = 3x² dx gives ln|y| = x³ + C, so y = Ae^(x³), and y(0) = 2 gives A = 2.

y⁢(x)=2ex3y⁢(0)=2

Keep exploring

Derivative & antiderivative checks opens with y = 3e^(x²) and verifies its derivative, 6xe^(x²), which is 2xy.

Return to the concept →
Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

Check the derivative Graph the solution Open worked example on a board Integral of 1/x in Math Reference

Your existing work stays on this device. Examples open as editable copies.