Calculus · real student question

Solve the differential equation dy/dx = ky, where k is a constant.

Question

Solve

dydx=ky\frac{dy}{dx} = ky

where kk is a constant.

Step-by-step solution

  1. Recognise the model. This equation says the rate of change is proportional to the current amount — the law behind population growth, radioactive decay, continuously compounded interest and Newtonian cooling. Solving it once solves all of them.

  2. Separate the variables. Dividing by yy requires y0y \ne 0, which we note and return to:

    dyy=kdx\frac{dy}{y} = k\,dx

  3. Integrate both sides.

    1ydy=kdxlny=kx+C0\int \frac{1}{y}\,dy = \int k\,dx \quad\Longrightarrow\quad \ln|y| = kx + C_0

    The absolute value is required because 1y\tfrac{1}{y} has antiderivative lny\ln|y| on either side of zero.

  4. Exponentiate and absorb the constants.

    y=ekx+C0=eC0ekx|y| = e^{kx + C_0} = e^{C_0}e^{kx}

    Writing A=eC0>0A = e^{C_0} > 0 and letting the sign of yy be carried by allowing C=±AC = \pm A, this becomes

    y=Cekxy = Ce^{kx}

  5. Recover the excluded solution. The division by yy ruled out y0y \equiv 0, but that constant function does satisfy the equation — and it is recovered by C=0C = 0. So allowing CC to be any real number makes y=Cekxy = Ce^{kx} the complete general solution with nothing lost.

  6. Verify and read off the behaviour. Differentiating, y=Ckekx=k(Cekx)=kyy' = Cke^{kx} = k\left(Ce^{kx}\right) = ky. With an initial condition y(0)=y0y(0) = y_0 we get C=y0C = y_0, so y=y0ekxy = y_0e^{kx}: growth when k>0k > 0, decay when k<0k < 0, and a constant when k=0k = 0. The doubling time for k>0k > 0 is ln2k\tfrac{\ln 2}{k}.

Answer

y=Cekx,CRy = Ce^{kx}, \qquad C \in \mathbb{R}

Need to solve a different problem like this? Open the solver →