# [SOLVED]1.5.7.19 IVP

#### karush

##### Well-known member
$\tiny{1.5.7.19}$
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
$ln(y)=-5t+ln(2)\implies ln\dfrac{y}{2}=-5t \implies \dfrac{y}{2}=e^{-5t} \implies y=2e^{-5t}$

ok this is my first pass thru this saw the $\dfrac{1}{y}y'$ idea in another example and tried to see if worked here but
the lesson has been to use $u(t)=exp(t)$ method but I couldn't resolve it...
typo's maybe

Last edited:

#### Cbarker1

##### Active member
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

##### Well-known member
6

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

#### Prove It

##### Well-known member
MHB Math Helper
Surely you mean $u(t) = \mathrm{e}^t$, not $\mathrm{e}^{\int{t}\,dt}$...

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