Solve the initial-value problem {y}''- 2{y}' = x + 2e^x, y(0) = 13, {y}'(0) = \frac{23}{4}.


Solve the initial-value problem {eq}{y}''- 2{y}' = x + 2e^x, y(0) = 13{/eq}, {eq}{y}'(0) = \frac{23}{4}{/eq}.

Solution of the Differential Equation:

The given differential equation is a second order linear differential equation. To find the solution of the differential equation first we find the auxiliary equation and then write the complementary solution in the form as follows

{eq}\displaystyle y_c=c_1e^{\alpha x}+c_2e^{\beta x} {/eq}

To find the particular solution we use the following property as follows

{eq}\displaystyle \left [ \frac{1}{f(D)}e^{ax}=\frac{1}{f(a)}e^{ax},\: \: f(a)\neq 0 \right ]\\ \displaystyle \left [ \frac{1}{f(D)}e^{ax}=\frac{x}{f{}'(D)}e^{ax},\: \: f(a= 0 \right ] {/eq}

Where {eq}a {/eq} is a constant and {eq}D\equiv \frac{\mathrm{d} }{\mathrm{d} x} {/eq}

Then the general solution of the differential equation will be in the form of

{eq}\displaystyle y(x)=y_c+y_p{/eq}

Answer and Explanation: 1

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Consider the differential equation

{eq}\displaystyle {y}''- 2{y}' = x + 2e^x,\: \: y(0) = 13,\quad y{}'(0) = \frac{23}{4} {/eq}

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First-Order Linear Differential Equations


Chapter 16 / Lesson 3

In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.

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