Question

Equations with the Dependent Variable Missing. For a scond order differential equation of the form \(y^{\prime \prime}=f\left(t, y^{\prime}\right),\) the substitution \(v=y^{\prime}, v^{\prime}=y^{\prime \prime}\) leads to a first order equation of the form \(v^{\prime}=f(t, v)\). If this equation can be solved for \(v\), then \(y\) can be obtained by integrating \(d y / d t=v .\) Note that one arbitrary constant is obtained in solving the first order equation for \(v,\) and a second is introduced in the integration for \(y\). In each of Problems 28 through 33 use this substitution to solve the given equation. $$ y^{\prime \prime}+y^{\prime}=e^{-t} $$

Short answer

To solve the given second-order differential equation, we first transformed it into a first-order differential equation by substituting v for y' and v' for y''. We then solved the first-order differential equation to find v and integrated v with respect to t to find y. The final solution is: $$ y(t) = -(t+C_1)e^{-t} - e^{-t} + C_2 $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Substituting v and v' for y' and y'' respectively

Given the second order differential equation: $$ y^{\prime\prime} + y^{\prime} = e^{-t} $$ Let's make the substitution \(v = y^{\prime}\), and \(v^{\prime} = y^{\prime\prime}\). The given equation becomes: $$ v^{\prime} + v = e^{-t} $$ Now we have a first-order differential equation in terms of \(v\) and \(t\).

Solve the first-order differential equation for v

The first-order differential equation in terms of \(v\) and \(t\) is: $$ v^{\prime} + v = e^{-t} $$ This is a first-order linear differential equation. To solve it, we need to find an integrating factor. The integrating factor \(I(t)\) for this equation is given by: $$ I(t) = \exp\left(\int 1 \, dt \right) = e^{t} $$ Now, multiply the equation by the integrating factor: $$ e^t v^{\prime} + e^t v = e^{-t} e^t $$ $$ \frac{d}{dt}\left(e^t v\right) = 1 $$ Now, integrate both sides with respect to t: $$ \int\frac{d}{dt}\left( e^t v\right) dt = \int 1 dt $$ $$ e^t v = t + C_1 $$ Now, isolate \(v\): $$ v = e^{-t}(t + C_1) $$

Integrate v to find y

Recall that \(v = y^{\prime}\). So now we need to integrate \(v\) with respect to \(t\) to find \(y\): $$ y^{\prime} = e^{-t}(t + C_1) $$ $$ y(t) = \int e^{-t}(t + C_1) dt $$ To integrate this expression, we can use integration by parts, let: $$ u = t + C_1,\ dv = e^{-t} dt \\ du = dt,\ v = -e^{-t} $$ Applying integration by parts formula (\(\int u dv = uv - \int v du\)), we get $$ y(t) = -(t+C_1)e^{-t} + \int e^{-t} dt $$ $$ y(t) = -(t+C_1)e^{-t} - e^{-t} + C_2 $$ Therefore, the solution to the original second-order differential equation is: $$ y(t) = -(t+C_1)e^{-t} - e^{-t} + C_2 $$ where \(C_1\) and \(C_2\) are arbitrary constants.