# Determine a unique solution of the separable differential

Determine a unique solution of the separable differential equa- tion that satisfies the given initial condition.
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Given Differential equation is
$\frac{{d}^{2}y}{{dt}^{2}}=\frac{dy}{dt}$
Boundary Condition is
$\frac{dy}{dt}\left(0\right)=2$ and $y\left(0\right)=3$
Solution:
$\frac{{d}^{2}y}{{dt}^{2}}=\frac{dy}{dt}$
Substitute $\frac{dy}{dt}={y}^{\prime }$
So,
$y{}^{″}={y}^{\prime }$
$y{}^{″}-{y}^{\prime }=0$
This is second order linear homogeneous ordinary differential equation
Auxillary equation is
${m}^{2}-m=0$
$m\left(m-1\right)=0$
Roots are$m=0$ and $m=1$
Roots are real and district
So solution is
$y\left(t\right)={C}_{1}{e}^{{m}_{1}t}+{C}_{2}{e}^{{m}_{2}t}$
Here put ${m}_{1}=0$ and ${m}_{2}=1$
So solution is
$y={C}_{1}{e}^{0\cdot t}+{C}_{2}{e}^{1\cdot t}$
$y\left(t\right)={C}_{1}+{C}_{2}{e}^{t}$
Use Given Initial Condition
$y\left(0\right)={C}_{1}+{C}_{2}{e}^{0}={C}_{1}+{C}_{2}=3$
$y\left(t\right)={C}_{1}+{C}_{2}{e}^{T}$
${y}^{\prime }\left(t\right)={C}_{2}{e}^{t}$
$y0\right)={C}_{2}=2$
$⇒{C}_{1}+2=3$
${C}_{1}=3-2=1$
$⇒{C}_{1}=1$
Put ${C}_{1}=1$ and ${C}_{2}$ in y(t)
$y\left(t\right)=1+2{e}^{t}$
$y\left(t\right)=1+2{e}^{t}$ is solution of given Differential Equation
Jeffrey Jordon