# Math Help - Differential Equation Help

1. ## Differential Equation Help

i need help on how to solve this differential equation

y'' = y' * e^y
initial conditions
y(0) = 0
y'(0) = 2

first i use u = y' and y'' = u*du/dy substitution to get
u*du/dy = u * e^y ---> du/dy = e^y

separating variables and integrating both sides give
u = e^y + c1

substituting y' back into u and separating variables give
dy / (e^y + c1) = dx

integrating gives
[ln(e^y) - ln(e^y + c1)] / c1 = x + c2
y - ln(e^y + c1) = c1*(x + c2)

from here i am stuck. I can't get it to be in the form of y =

2. Originally Posted by TshingY
i need help on how to solve this differential equation

y'' = y' * e^y
initial conditions
y(0) = 0
y'(0) = 2

first i use u = y' and y'' = u*du/dy substitution to get
u*du/dy = u * e^y ---> du/dy = e^y

separating variables and integrating both sides give
u = e^y + c1 (*)

substituting y' back into u and separating variables give
dy / (e^y + c1) = dx

integrating gives
[ln(e^y) - ln(e^y + c1)] / c1 = x + c2
y - ln(e^y + c1) = c1*(x + c2) (**)

from here i am stuck. I can't get it to be in the form of y =
Why not use your IC here (*) to find the constant c1 (=1). Second, leave the ln alone

$\ln e^y - \ln (e^y + 1) = \ln \frac{e^y}{e^y+1} = \cdots$ so you can solve for $e^y$ then take a ln to get y.

3. If $y''=\mathrm e^yy'$ then integration gives $y'=\mathrm e^y+c$.

Since $y(0)=0$ and $y'(0)=2$ we see that $c=1$.

Thus $\frac{\mathrm d y}{\mathrm e^y+1}=\mathrm d x$.

Write this as $\frac{\mathrm e^{-y}\mathrm d y}{1+\mathrm e^{-y}}=\mathrm d x$.

Now you may integrate to get $-\ln(1+\mathrm e^{-y})=x+k$.

Since $y=0$ when $x=0$ this means $k=-\ln 2$.

Hence $x=\ln\left(\frac 2{1+\mathrm e^{-y}}\right)$ and so $\mathrm e^x=\frac 2{1+\mathrm e^{-y}}$.

Rearrange to get $y=-\ln(2\mathrm e^{-x}-1)$.