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
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
$\displaystyle -e^{-y}=-\frac{1}{2}e^{-x^2}+c$
$\displaystyle e^{-y}=\frac{1}{2}e^{-x^2}+c$
$\displaystyle \ln(e^{y})=\ln(\frac{1}{2}e^{-x^2}+c)$
$\displaystyle y=\ln(\frac{1}{2}e^{-x^2}+c)$
Now using the IC
$\displaystyle 0=\ln(\frac{1}{2}e^{0}+c) \iff 0=\ln(\frac{1}{2}+c) \iff 1=\frac{1}{2}+c \iff \frac{1}{2}=c$