# Find the unique solution of the second-order initial value problem. y''+14y=0, y(0)=-1,y'(0)=2

## Question:

Find the unique solution of the second-order initial value problem.

{eq}y''+14y=0, y(0)=-1,y'(0)=2 {/eq}

## Initial Value Problem:

Differential equation of form {eq}\displaystyle \:\phi \left(D\right)y=f\left(x\right) {/eq} has solution {eq}\displaystyle y=y_c+y_p {/eq},

Where,{eq}\displaystyle y_c {/eq} is called complementary function and {eq}\displaystyle y_p {/eq} is called particular integral.

Use initial condition to find the value of constants.

## Answer and Explanation: 1

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Given differential equation is

{eq}\displaystyle y''+14y=0, y(0)=-1,y'(0)=2 {/eq}

==> {eq}\displaystyle...

See full answer below.

#### Learn more about this topic: First-Order Linear Differential Equations

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Chapter 16 / Lesson 3
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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.