## Related questions with answers

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

Solution

VerifiedIt is given that:

$\begin{align*} y^{\prime\prime}-2y^\prime-3y&=0\\ y(0)&=2\\ y^\prime(0)&=2 \end{align*}$

In order to solve the initial-value problem we will first solve the given differential equation.,The auxiliary equation is:

$r^2-2r-3=0$

Now,let us solve this quadratic equation:

$\begin{align*} r_{1,2}&=\dfrac{2 \pm \sqrt{(-2)^2-4\cdot 1 \cdot (-3)}}{2}\\ &=\dfrac{2\pm \sqrt{4+12}}{2}\\ &=\dfrac{2\pm \sqrt{16}}{2}\\ &=\dfrac{2\pm 4}{2}\Rightarrow \pmb{r_1=3\,\, , \,\, r_2=-1} \end{align*}$

Notice that this is the case when $b^2-4ac>0$ so the general solution has the form

$y(x)=c_1 e^{r_1x}+c_2 e^{r_2x}$

Substituting $r_1$ and $r_2$ we conclude that the general solution of the given differential equation is

$\boxed{y(x)=c_1 e^{3x}+c_2 e^{-x}}$

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