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Solve the differential equations. \(y^{\prime \prime}+25 y=0\)

Short Answer

Expert verified
The general solution of the differential equation \( y^{\prime \prime}+25 y=0 \) is \( y = A cos(5t)+ B sin(5t) \)

Step by step solution

01

Formulate the Characteristic Equation

The characteristic equation is obtained from the coefficients of the differential equation. For this equation, it becomes: \( r^2 + 25 = 0 \)
02

Find the Roots of the Characteristic Equation

Let's solve the characteristic equation we just formulated. Subtracting 25 from both sides we get \( r^2 = -25 \). Taking the square root of both sides leads to \( r = ±5i \). These are complex roots where real part is 0 and imaginary part is 5.
03

Write the General Solution

Given a second order homogeneous differential equation with complex roots \( a ± bi \), the general solution is: \( y = e^{at}(A cos(bt)+ B sin(bt)) \). In this case, where the real part is 0, the factors involving \( e^{at} \) become 1. Therefore, the general solution is : \( y = A cos(5t)+ B sin(5t) \). A and B are arbitrary constants that can be determined by initial conditions if such are provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

The Characteristic Equation in Differential Equations
The characteristic equation is a crucial part of solving linear differential equations. It's like a bridge that connects the differential equation to its solution. In the case of a second-order differential equation such as \(y'' + 25y = 0\), the characteristic equation is derived by replacing the differential operator with a hypothetical polynomial. For this specific equation, we assume a solution of the form \(e^{rt}\), where \(r\) is a constant, which transforms the differential equation into \(r^2 + 25 = 0\).

This characteristic equation is essentially a polynomial equation that allows us to find the roots which determine the behavior of the solution. Setting it up correctly is the first crucial step in solving the differential equation. It indicates whether the solution will include exponential, oscillatory, or a combination of behaviors by the nature of its roots, leading us toward the general solution.
Understanding Complex Roots
Complex roots appear when solving the characteristic equation leads to solutions with negative discriminants. For the equation \(r^2 + 25 = 0\), solving it involves taking the square root of a negative number, which results in complex numbers. Here, you get roots that are \(r = \pm 5i\). These roots have no real component and include an imaginary number, signified by the presence of \(i\), where \(i^2 = -1\).

In general, complex roots are of the form \(a \pm bi\), although in this particular problem, the real part \(a\) is zero, and the imaginary part \(b\) is 5. Complex roots introduce trigonometric functions into the solution. This oscillatory behavior reflects the sine and cosine components of the solution, indicating the recurring or wave-like nature of the solutions relative to time or space.
Finding the General Solution
The general solution gives us the complete set of possible solutions to the differential equation, encapsulating the behavior of the system described. With complex roots, as found from our characteristic equation \(r = \pm 5i\), the general solution involves trigonometric functions. The general form is given by \[y(t) = e^{at}(A\cos(bt) + B\sin(bt))\], where \(a\) is the real part and \(b\) is the imaginary part of the roots.

When the real part \(a\) is zero, \(e^{at}\) simplifies to 1, thus eliminating exponential growth or decay in the solution. Therefore, the solution becomes \(y(t) = A\cos(5t) + B\sin(5t)\), revealing purely oscillatory behavior. Here, \(A\) and \(B\) are arbitrary constants determined by initial or boundary conditions. This structure of the general solution demonstrates how varying combinations of sine and cosine functions can describe a range of behaviors for the original differential equation.

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Most popular questions from this chapter

A 5-gallon urn is filled with chai, a milky spicy tea. The chai in the urn is \(90 \%\) tea and \(10 \%\) milk. Chai is being consumed at a rate of \(1 / 2\) gallon per hour and the urn is kept full by adding a mixture that is \(80 \%\) tea and \(20 \%\) milk. Assume that the chai is well-mixed. (a) Write a differential equation whose solution is \(M(t)\), the number of gallons of milk in the urn at time \(t\). Specify the initial condition. (b) Use qualitative analysis to sketch the solution to the differential equation in part (a). (c) How much milk is in the urn after 2 hours?

Solve the differential equations below. Find the general solution. (a) \(\frac{d y}{d t}=3 t+5\) (b) \(\frac{d y}{d t}=3 y\) (c) \(\frac{d y}{d t}=-y\) (d) \(\frac{d y}{d t}=0\) (e) \(\frac{d y}{d t}=3 y-6\)

For what value(s) of \(\beta\), if any, is (a) \(y=C_{1} \sin \beta t\) a solution to \(y^{\prime \prime}=-16 y ?\) (b) \(y=C_{2} \cos \beta t\) a solution to \(y^{\prime \prime}=-16 y\) ? (c) \(y=C_{3} e^{\beta t}\) a solution to \(y^{\prime \prime}=-16 y\) ?

Let \(P(t)\) be the number of crocodiles in a mud hole at time \(t\). Suppose \(\frac{d P}{d t}=0.01 P-0.0025 P^{2}\) (a) What is the carrying capacity of the mud hole? (b) Find \(\frac{d^{2} P}{d t^{2}}\). Remember: You are differentiating with respect to \(t\), so the derivative of \(P\) is not 1 . (c) Use your answer to part (b) to determine how many crocodiles are in the mud hole when the number of crocodiles is increasing most rapidly. (d) Sketch a solution curve if the number of crocodiles in the mud hole at time \(t=0\) is 3. (Label the vertical axis. You need not calibrate the \(t\) -axis.)

Which of the following is a solution to the differential equation $$ y^{\prime \prime}+9 y=0 ? $$ (a) \(y=e^{3 t}+e^{-3 t}\) (b) \(y=C e^{t}-t\) (c) \(y=C\left(t^{2}+t\right)\) (d) \(y=\sin 3 t+6\) (e) \(y=5 \cos 3 t\)

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