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.

Key concepts

Solution Verification

Solution verification is the process of confirming that a proposed solution satisfies a given differential equation. To achieve this, we substitute the function and its derivatives back into the equation. If the equation holds true (i.e., both sides are equal), the function is verified as a solution.
In this exercise, we have two functions, \(y_1(t)\) and \(y_2(t)\), which need verification against the differential equation \[ y^{m}+4y'' +3y = t.\]

• For each function, we compute its derivatives.
• Substitute the function and its derivatives into the equation.
• Check if the left-hand side equals the right-hand side \(t\).

By following these steps, we confirm that both functions \(y_1(t)=t/3\) and \(y_2(t)=e^{-t}+t/3\) satisfy the differential equation, proving them as valid solutions.

First Derivative

The first derivative provides the rate at which a function changes with respect to a variable, commonly time \(t\). Calculating the first derivative is essential in verifying solutions to differential equations.
For the function \(y_1(t) = t/3\), the first derivative is straightforward:\[ y_1'(t) = \frac{d(t/3)}{dt} = \frac{1}{3}. \]
Similarly, for \(y_2(t) = e^{-t} + t/3\), the derivative is:\[ y_2'(t) = \frac{d(e^{-t} + t/3)}{dt} = -e^{-t} + \frac{1}{3}. \]

• The derivative of \(t/3\) simplifies to a constant.
• The derivative of \(e^{-t}\) results in \(-e^{-t}\), indicating exponential decay.

Understanding these derivatives allows us to test the behavior of the functions in the differential equation.

Second Derivative

The second derivative provides information on the curvature or acceleration of a function, indicating how the rate of change of the function itself changes.

• For \(y_1(t) = t/3\), the second derivative is:\[ y_1''(t) = \frac{d^2(t/3)}{dt^2} = 0. \]
• This zero second derivative indicates a linear graph without curvature.
• For \(y_2(t) = e^{-t} + t/3\), the second derivative becomes:\[ y_2''(t) = \frac{d^2(e^{-t} + t/3)}{dt^2} = e^{-t}. \]
• This positive second derivative shows a convex curvature due to the exponential function.

These derivatives play a pivotal role in substitution into the original differential equation to verify the solution.

Substitution Method

The substitution method involves entering the original function and its derivatives into the differential equation to check for an identity. This verifies whether the solution truly satisfies the equation.
For our case:

• We substitute \(y_1(t)\) with its derivatives into \[ y^{m}+4y''+3y. \]
• Calculate: \[ y_1: 0 + 4(0) + 3(t/3) = t. \]
• Repeat for \(y_2(t)\) with its derivatives: \[ y_2: e^{-t} + 4(e^{-t}) + 3(e^{-t} + t/3) = t. \]

This step confirms that both \(y_1(t)\) and \(y_2(t)\) satisfy the differential equation by achieving identity on both sides, thus being valid solutions.