## Related questions with answers

Solve the initial value problem. $y^{\prime \prime}+2 y^{\prime}-3 y=0 ; y(0)=6, y^{\prime}(0)=-2$

Solutions

Verifiedwe solve the associated characteristic equation for the given initial value problem.

$\begin{gather*} \lambda^2+2\lambda-3=(\lambda+3)(\lambda-1)=0\\ \Rightarrow \lambda_{1}=1, \lambda_{2}=-3\\ \text{Because we have real roots the general solution is associated with}\\ y=c_{1}e^x+c_{2}e^{-3x} \end{gather*}$

Consider the initial value problem $y''+2y'-3y=0$; $y(0)=6$, $y'(0)=-2$.

The corresponding characteristic equation is $\lambda^2+2\lambda-3=0$.

Since $\lambda^2+2\lambda-3=\left(\lambda+3\right)\left(\lambda-1\right)$, we have that the characteristic equation has $-3$ and $1$ as its roots.

Thus, the general solution to the differential equation is $y=c_1e^{-3x}+c_2e^{x}$ which gives us that $y'=-3c_1e^{-3x}+c_2e^x$.

Since $y(0)=6$ and $y'(0)=-2$ we have that $c_1+c_2=6$ and $-3c_1+c_2=-2$ which gives us that $c_1+c_2-\left(-3c_1+c_2\right)=6-(-2)$ and therefore, $4c_1=8$.

Hence, we have that $c_1=2$ and then since $c_1+c_2=6$ we have that $c_2=4$.

Thus, the solution to the initial value problem is $y=2e^{-3x}+4e^x$.

## Create an account to view solutions

## Create an account to view solutions

## Recommended textbook solutions

#### Advanced Engineering Mathematics

10th Edition•ISBN: 9780470458365 (4 more)Erwin Kreyszig#### Advanced Engineering Mathematics

9th Edition•ISBN: 9780471488859 (1 more)Erwin Kreyszig#### Advanced Engineering Mathematics

7th Edition•ISBN: 9781111427412 (2 more)Peter V. O'Neil#### Advanced Engineering Mathematics

6th Edition•ISBN: 9781284105902 (8 more)Dennis G. Zill## More related questions

1/4

1/7