Solve the differential equation: y'-2xy=0. Write answer in the form y=f(x).

Question

Solve the differential equation: y'-2xy=0. Write answer in the form y=f(x).

Answer 1

This is a linear first order differential equation

Integrating factor = e^∫(-2x)dx = e^{-x^2}

Multiply both sides by the integrating factor to get y'e^{-x^2} - 2xe^{-x^2}y = 0

By the Product rule, we see that the left hand side is the derivative of ye^{-x^2}

So we have (ye^{-x^2})' = 0

Integrate to obtain ye^{-x^2} = C

Thus, y = Ce^{x^2}, where C is an arbitrary constant.

Answer 2

We can use separation of variables to find y=f(x).

y'-2xy = 0

y' = 2xy

y'/y = 2x

Integrate both side

ln(y) = x² + c1

e^ln(y) = y = e^{x^2+c1} = e^{x^2}e^c1=Ce^{x^2}

y = Ce^{x^2}

If we had initial condition, we could solve for C.