- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

$\tiny{1.5.7.19}$

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

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

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