- #1

karush

Gold Member

MHB

- 3,269

- 5

$\tiny{1.5.7.19}$

\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}$

\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}$

Last edited: