Solving a Differential Equation

[karush]

$\tiny{1.5.7.19}$
\nmh{157}
Solve the initial value problem
$y'+5y=0\quad y(0)=2$
$u(x)=exp(5)=e^{5t+c_1}$?

so tried
$\dfrac{1}{y}y'=-5$
$ln(y)=-5t+c_1$

apply initial values
$ln(y)=-5t+ln(2)\implies ln\dfrac{y}{2}=-5t
\implies \dfrac{y}{2}=e^{-5t}
\implies y=2e^{-5t}$

 

[cbarker1]

We will return to your problem after a short review of what input of a function means. Let's suppose that $f(x)=x+3$. What is $f(3)$?

 

[karush]

6

here $u(t)=exp(t)= e^{\int t \ dt}$

 

[Prove It]

Surely you mean $u(t) = \mathrm{e}^t$, not $\mathrm{e}^{\int{t}\,dt}$...

In actuality, the integrating factor is $u(t) = \mathrm{e}^{5\,t}$...