can anyone help me out wif this problem??
maybe...its an easy cake for u..
dy/dx=xe^(y-x^2) and y(0)=0
I have to solve the initial value....
and show all intermediate steps in my solution of the differential equation.
thanks
The LHS is simply -e^(-y)
for the RHS use the u substitution u = -x^2
you obtain -e^(-y) = -1/2 e^(-x^2) + C
e^(-y) = 1/2 e^(-x^2) + C
since y(0) = 0
1 = 1/2 + C C =1/2
e^(-y) = 1/2 e^(-x^2) + 1/2
now use logarithms to solve for y-- I/ll let you finish that part
int(e^-y)dy=int(xe^(-x^2))dx,
Let u=-x^2
du/dx=-2x
dx=du/-2x
then, it becomes,
int(e^-y)dy=int(-e^u)du
-------------------------
this is the part that confuses me!!!
would it become...
-e^-y= -e^u +C
-e^-y=-e^(-x^2) +C
C= -e^-y + e^(-x^2)
given that, y(0)=0,
substitute y=0 and x=0 to find C,
then,
C=-1 + 1=0
is this right???
I get all your explanations thanks a lot!!!
but.. shouldn't C be equal to -1/2
-e^(-y) = -1/2 e^(-x^2) + C
since, y(0)=0
then,
doesn't it become:
-1=-1/2 +C
C= -1/2
Thus, the expression is now then,
-e^(-y) = -1/2 e^(-x^2) -1/2????
and also not sure how to solve y----.. T_T using logarithm
thanks you so much
i have to solve y---, solving the initial value
given, dy/dx=xe^(y-x^2) and y(0)=0
using differential equation,
I hav done it so far up to this step...
------------------------------
-e^(-y) = -1/2 e^(-x^2) + C
since, y(0)=0
then,
doesn't it become:
-1=-1/2 +C
C= -1/2
Thus, the expression is now then,
-e^(-y) = -1/2 e^(-x^2) -1/2????
and also not sure how to solve y----.. T_T using logarithm for the next step..
thanks you so muc