17Calculus - Differential Equations - Particular Solutions

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When we solve differential equations, we will usually use integration. As you remember from integration, if you don't have initial conditions, you end up with a generic, unknown constant in your answer as a placeholder indicating that there are actually infinite solutions. The same concept appears when solving differential equations.

General vs Particular Solution

What is the difference between the general solution and the particular solution?

Short Answer - - The terminology relates to whether or not we have an unknown constant in our final answer. For the general solution, an unknown constant IS part of the solution indicating an infinite number of solutions. In the particular solution, there is no unknown constant and the solution we have is the one and only solution.

Long Answer - - When you solve a differential equation, you use integration, which introduces an unknown constant.
In the general solution, the unknown constant remains and you do not have enough information to be able to determine what that constant is. Consequently, you have an infinite number of solutions.
In the particular solution, you start out by solving for the general solution but then you are given initial conditions which you use to determine the value of the constant. These initial conditions are actually points that the solution to the differential equation pass through. In the end, you have only one solution without any unknown constants.

general solution particular solution infinite number of solutions only one solution contains unknown constant(s) does NOT contain any unknown constants no (or too few) initial conditions given initial conditions given; used to solve for constants

To find the general solution, just solve the differential equation and leave any unknown constants in your final answer.
To find the particular solution, find the general solution first, then plug in the initial conditions and solve for the constants.

Notes
1. When determining the particular solution, you will be given the same number of initial conditions as the order of the differential equation. Depending on how the initial conditions are given, you may need to stop after each integration and solve for the individual constants or you may need to wait until you are completely done and solve for all the constants at once. You will get a feel for this as you work practice problems.
2. For a general solution, if no initial conditions are given or fewer than the order of the differential equation are given, we cannot determine all of the unknown constants, since each integration introduces another constant.
3. When you were first learning integration, you probably ran across initial value problems. These were actually differential equations where you were asked to find the particular solution.

Practice

Verify that $$y = x^3 - c/x$$ is a solution to $$2xy' + 2y = 8x^3$$ and find the value of $$c$$ for which the solution satisfies the initial condition $$y(2) = 6$$.

Problem Statement

Verify that $$y = x^3 - c/x$$ is a solution to $$2xy' + 2y = 8x^3$$ and find the value of $$c$$ for which the solution satisfies the initial condition $$y(2) = 6$$.

Solution

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Verify that $$y=c_1 \sin(2x) + c_2 \cos(2x) + 1$$ is the general solution to $$y'' + 4y = 8$$ and solve for $$c_1$$ and $$c_2$$ if $$y(0) = 5$$ and $$y'(0) = -1$$.

Problem Statement

Verify that $$y=c_1 \sin(2x) + c_2 \cos(2x) + 1$$ is the general solution to $$y'' + 4y = 8$$ and solve for $$c_1$$ and $$c_2$$ if $$y(0) = 5$$ and $$y'(0) = -1$$.

Solution

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Verify that $$y=Ce^x - 3x - 3$$ is the general solution to $$dy/dx = 3x + y$$ and solve for $$C$$ if $$y(0) = -1$$.

Problem Statement

Verify that $$y=Ce^x - 3x - 3$$ is the general solution to $$dy/dx = 3x + y$$ and solve for $$C$$ if $$y(0) = -1$$.

Solution

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Verify that $$y=x^2 + c/x^2$$ is the general solution to $$xy' + 2y = 4x^2$$ and solve for $$c$$ if $$y(5) = 8$$.

Problem Statement

Verify that $$y=x^2 + c/x^2$$ is the general solution to $$xy' + 2y = 4x^2$$ and solve for $$c$$ if $$y(5) = 8$$.

Solution

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The general solution of a differential equation is $$y = ae^{4t} + be^{6t}$$ where $$a$$ and $$b$$ are constants. Use the initial conditions $$y(0) = 2$$ and $$y'(0) = 5$$ to find the particular solution.

Problem Statement

The general solution of a differential equation is $$y = ae^{4t} + be^{6t}$$ where $$a$$ and $$b$$ are constants. Use the initial conditions $$y(0) = 2$$ and $$y'(0) = 5$$ to find the particular solution.

Solution

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Verify that $$y = ce^{-2x} + e^{-x}$$ is a solution to the differential equation $$y' + 2y = e^{-x}$$ and determine the value of $$c$$ for which the solution satisfies the initial condition $$y(-4) = 4$$.

Problem Statement

Verify that $$y = ce^{-2x} + e^{-x}$$ is a solution to the differential equation $$y' + 2y = e^{-x}$$ and determine the value of $$c$$ for which the solution satisfies the initial condition $$y(-4) = 4$$.

Solution

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