Consider The Following IVP: U"(t) + U'(t) - 12u(t)=0 (1) U(0) = 70 And U'(0) = 18. Show That U(t)=ce"

Consider The Following IVP: U"(t) + U'(t) - 12u(t)=0 (1) U(0) = 70 And U'(0) = 18. Show That U(t)=ce"

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Introduction to the Initial Value Problem (IVP)

In the realm of differential equations, initial value problems (IVPs) serve as foundational tools for modeling real-world phenomena across physics, engineering, and other scientific disciplines. The specific IVP under consideration involves a second-order linear differential equation:

\[ U''(t) + U'(t) - 12 U(t) = 0 \]

along with the initial conditions:

\[ U(0) = 70 \quad \text{and} \quad U'(0) = 18 \]

The goal is to demonstrate that the solution \(U(t)\) can be expressed in the form:

\[ U(t) = c e^{\lambda t} \]

where \(c\) and \(\lambda\) are constants to be determined. This process involves solving the differential equation systematically and applying the initial conditions to find the specific solution.

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Understanding the Differential Equation

Type of Differential Equation

The given differential equation:

\[ U''(t) + U'(t) - 12 U(t) = 0 \]

is a homogeneous linear second-order differential equation with constant coefficients. Such equations are prevalent in modeling oscillations, exponential growth/decay, and other dynamic systems.

Significance of the Equation

• Homogeneity: The right-hand side is zero, indicating no external forcing.
• Linearity: The superposition principle applies, meaning solutions can be added.
• Constant coefficients: Simplifies the solution process via characteristic equations.

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Solving the Differential Equation

Step 1: Write the Characteristic Equation

To solve the differential equation, assume a solution of the form:

\[ U(t) = e^{\lambda t} \]

Substituting into the differential equation yields:

\[ \lambda^2 e^{\lambda t} + \lambda e^{\lambda t} - 12 e^{\lambda t} = 0 \]

Dividing through by \(e^{\lambda t}\) (which is never zero):

\[ \lambda^2 + \lambda - 12 = 0 \]

This quadratic equation, known as the characteristic equation, determines the form of the general solution.

Step 2: Find the Roots of the Characteristic Equation

Solve:

\[ \lambda^2 + \lambda - 12 = 0 \]

Using the quadratic formula:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \(a=1\), \(b=1\), and \(c=-12\):

\[ \lambda = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-12)}}{2} \]
\[ \lambda = \frac{-1 \pm \sqrt{1 + 48}}{2} \]
\[ \lambda = \frac{-1 \pm \sqrt{49}}{2} \]
\[ \lambda = \frac{-1 \pm 7}{2} \]

Thus, the roots are:

• \( \lambda_1 = \frac{-1 + 7}{2} = 3 \)


• \( \lambda_2 = \frac{-1 - 7}{2} = -4 \)

Step 3: Write the General Solution

Since the roots are real and distinct, the general solution is:

\[ U(t) = C1 e^{\lambda1 t} + C2 e^{\lambda2 t} \]
\[ U(t) = C1 e^{3t} + C2 e^{-4t} \]

where \(C1\) and \(C2\) are arbitrary constants determined by initial conditions.

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Applying Initial Conditions to Find Specific Solution

Step 4: Compute \(U'(t)\)

Differentiate \(U(t)\):

\[ U'(t) = 3 C1 e^{3t} - 4 C2 e^{-4t} \]

Step 5: Use Initial Conditions

Given:

\[ U(0) = 70 \]
\[ U'(0) = 18 \]

Plugging in \(t=0\):

\[ U(0) = C1 e^{0} + C2 e^{0} = C1 + C2 \]
\[ 70 = C1 + C2 \quad \Rightarrow \quad (1) \]

Similarly, for \(U'(0)\):

\[ U'(0) = 3 C1 e^{0} - 4 C2 e^{0} = 3 C1 - 4 C2 \]
\[ 18 = 3 C1 - 4 C2 \quad \Rightarrow \quad (2) \]

Step 6: Solve for \(C1\) and \(C2\)

From equation (1):

\[ C2 = 70 - C1 \]

Substitute into equation (2):

\[ 18 = 3 C1 - 4(70 - C1) \]
\[ 18 = 3 C1 - 280 + 4 C1 \]
\[ 18 = 7 C_1 - 280 \]

Add 280 to both sides:

\[ 298 = 7 C_1 \]
\[ C_1 = \frac{298}{7} \approx 42.57 \]

Now, find \(C_2\):

\[ C_2 = 70 - \frac{298}{7} = \frac{490}{7} - \frac{298}{7} = \frac{192}{7} \approx 27.43 \]

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Final Solution and Expression

The specific solution to the initial value problem is:

\[ U(t) = \frac{298}{7} e^{3t} + \frac{192}{7} e^{-4t} \]

which can be written as:

\[ U(t) = C1 e^{3t} + C2 e^{-4t} \]

with explicit constants:

• \( C_1 = \frac{298}{7} \)


• \( C_2 = \frac{192}{7} \)

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Showing the Solution in the Form \( U(t) = c e^{\lambda t} \)

While the general solution involves two exponential terms, the statement "Show that \( U(t) = c e^{\lambda t} \)" suggests examining particular solutions or specific cases where the solution reduces to a single exponential form.

Case 1: When the roots are equal

This occurs if the characteristic equation has a repeated root:

\[ \lambda^2 + \lambda - 12 = 0 \]

has discriminant:

\[ \Delta = 1^2 - 4 \times 1 \times (-12) = 1 + 48 = 49 \neq 0 \]

Hence, roots are distinct, and the solution is a linear combination of two exponentials, not a single exponential.

Case 2: When the solution simplifies to a single exponential

Suppose we consider specific initial conditions that lead to one of the constants being zero:

• If \( C2 = 0 \), then \( U(t) = C1 e^{3t} \)


• If \( C1 = 0 \), then \( U(t) = C2 e^{-4t} \)

In the context of the initial conditions, this is not the case, as both constants are non-zero.

Conclusion

Therefore, the general solution involves a linear combination of two exponential functions, and the specific solution determined by initial conditions does not reduce to a single exponential of the form \( c e^{\lambda t} \). However, in theoretical contexts, particular solutions of the form \( U(t) = c e^{\lambda t} \) are fundamental, especially when considering homogeneous equations with a single characteristic root or when analyzing particular modes.

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Summary and Final Remarks

• The differential equation \( U''(t) + U'(t) - 12 U(t) = 0 \) is a second-order linear homogeneous differential equation with constant coefficients.
• The characteristic equation yields roots \( \lambda = 3 \) and \( \lambda = -4 \).
• The general solution is a linear combination:

\[ U(t) = C1 e^{3t} + C2 e^{-4t} \]

• Applying initial conditions \( U(0) = 70 \) and \( U'(0) = 18 \), we find the specific constants:

\[ C1 = \frac{298}{7}, \quad C2 = \frac{192}{7} \]

• The solution can be expressed explicitly as:

\[ U(t) = \frac{298}{7} e^{3t} +