Question

Solve the following initial value problem for u as a function of t:du/dt + (k/m) u = 0 (k and m positive constants), u(0) = u₀a. as a first-order linear equation.

Accepted Answer

Recognize that the given differential equation is a first-order linear ordinary differential equation of the form $$\frac{du}{dt} + \frac{k}{m} u = 0$$, where $$k$$ and $$m$$ are positive constants. Identify the integrating factor $$\mu(t)$$, which is given by $$\mu(t) = e$$^{\int P(t) \, dt}$$, where P(t$$)$$ is the coefficient of u$ in the equation. Here, P(t$$) = \frac{k}{m}$$, so calculate $$\mu(t) = $$e^{\int \frac{k}$${m} \, dt} = $$e^{\frac{k}$${m} t}$$. Multiply both sides of the differential equation by the integrating factor $$\mu(t)$$ to obtain e$$^{\frac{k}$${m} t} \frac{du}{dt} + e$$^{\frac{k}$${m} t} \frac{k}{m} u = 0. $$This simplifies to $$\frac{d}{dt} \left( e$$^{\frac{k}$${m} t} u \right) = 0$$ because the left side is the derivative of the product $$e^{\frac{k}$${m} t} u$$. Integrate both sides with respect to t$ to get e$$^{\frac{k}$${m} t} u = C$$, where $$C is $$the constant of integration. Solve for $$u(t) by $$dividing both sides by $$e^{\frac{k}$${m} t}$$, yielding u(t) = C$$ $$e^{-\frac{k}$${m} t}$$. Then, apply the initial condition u(0$$) = $$u_0 to $$find $$C = u_0$, so the solution is u(t$$) = $$u_0$$ $$e^{-\frac{k}$${m} t}$$.

Solve the following initial value problem for u as a function of t:

du/dt + (k/m) u = 0 (k and m positive constants), u(0) = u₀

a. as a first-order linear equation.

Verified video answer for a similar problem:

Key Concepts

First-Order Linear Differential Equations

Integrating Factor Method

Initial Value Problems