Question

Verify that the given function or functions is a solution of the differential equation. $$ y^{m}+4 y^{\prime \prime}+3 y=t ; \quad y_{1}(t)=t / 3, \quad y_{2}(t)=e^{-t}+t / 3 $$

Short answer

Answer: Yes, both functions \(y_{1}(t)=t / 3\) and \(y_{2}(t)=e^{-t}+t / 3\) are solutions to the given differential equation.

Step-by-step solution

Compute the first derivatives of \(y_{1}(t)\) and \(y_{2}(t)\)

For \(y_{1}(t)=t / 3\), the first derivative \(y_{1}'(t)\) can be computed as: $$ y_{1}^{\prime}(t)=\frac{d(t / 3)}{dt}=\frac{1}{3} $$ For \(y_{2}(t)=e^{-t}+t / 3\), the first derivative \(y_{2}'(t)\) can be computed as: $$ y_{2}^{\prime}(t)=\frac{d(e^{-t}+t / 3)}{dt}= -e^{-t}+\frac{1}{3} $$

Compute the second derivatives of \(y_{1}(t)\) and \(y_{2}(t)\)

For \(y_{1}(t)=t / 3\), the second derivative \(y_{1}^{\prime \prime}(t)\) can be computed as: $$ y_{1}^{\prime \prime}(t)=\frac{d^2(t / 3)}{dt^2}=0 $$ For \(y_{2}(t)=e^{-t}+t/3\), the second derivative \(y_{2}^{\prime \prime}(t)\) can be computed as: $$ y_{2}^{\prime \prime}(t)=\frac{d^2(e^{-t}+t/3)}{dt^2}=e^{-t} $$

Substitute the functions and their derivatives into the differential equation

Now we will substitute the functions and their derivatives into the differential equation \(y^{m}+4 y^{\prime \prime}+3 y=t\) and check if the left-hand side of the equation equals the right-hand side. For \(y_{1}(t)=t / 3\): $$ \begin{aligned} y^{m}+4 y^{\prime \prime}+3 y &= t^{1}+4(0)+3\left(\frac{t}{3}\right) \\ &= t+0+\frac{3t}{3} \\ &= t \end{aligned} $$ For \(y_{2}(t)=e^{-t}+t / 3\): $$ \begin{aligned} y^{m}+4 y^{\prime \prime}+3 y &= t^{e ^{-t}}+4 e^{-t}+3(e^{-t}+\frac{t}{3}) \\ &= t\cdot e^{-t}+4 e^{-t}+3e^{-t}+t \\ &= t \end{aligned} $$ Both left-hand sides of the equations equal the right-hand side \(t\). Thus, the given functions \(y_{1}(t)\) and \(y_{2}(t)\) are indeed solutions to the differential equation.