<p>i need to find the integral of y' = 1+y
dy/dx = y+1
dy/y+1 = dx
ln(y+1) = x
y+1=e^x
y= -1 + e^x</p>
<p>However the correct answer should be y= -1 + 2e^x</p>
<p>what did i do wrong?</p>
<p>i believe it has to do with ln()
i think ln()=2e
and log()=e</p>
<p>ln(y+1) = x<br>
y+1=e^x should be 2e because its a ln()</p>
<p>The correct answer is y=ce^x - 1 for any constant c</p>
<p>dy/dx = y+1
dy/y+1 = dx
ln(y+1) = x + c
y+1=e^(x+c)
y+1=e^c * e^x
y+1=ce^x
y=ce^x -1</p>
<p>You could only say that y=2e^x-1 if you were given some initial condition such as y(1) = 2e-1</p>
<p>(The original problem is a 1st order differential equation, and any y=ce^x - 1 solves it. You can check this by calculating y and y' (which =ce^x) and plugging them into the orignial equation)</p>
<p>tanman you correct, i forget the conditon which was y(0)=1 , so now i get the correct answer, thks</p>