# Thread: can anyone help me out solving differential equa.?? plz

1. ## can anyone help me out solving differential equa.?? plz

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

2. Originally Posted by mookie
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 ODE separates

$
\int e^{-y}\,dy = \int x e^{-x^2}\,dx + c
$

3. rewrite

dy/dx = x e^y e^(-x^2)

Separate

e^(-y) dy = x e^(-x^2) dx

Now you have 2 simple integrals so integrate

Finally apply the initial condition to determine C

4. ## integration..

can u plz how me nxt steps??
integrations

Originally Posted by danny arrigo
The ODE separates

$
\int e^{-y}\,dy = \int x e^{-x^2}\,dx + c
$

5. 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

6. ## from here...

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???

Originally Posted by Calculus26
rewrite

dy/dx = x e^y e^(-x^2)

Separate

e^(-y) dy = x e^(-x^2) dx

Now you have 2 simple integrals so integrate

Finally apply the initial condition to determine C

7. not quite

int(e^-y)dy=int(-e^u)du

should be

int(e^-y)dy=-1/2int(-e^u)du

remember -du/2 = xdx

8. thx!!!so much

so, i got

-e^(-y)= 1/2 e^(-x^2)+1/2...

how do u use logarithm to find out y

lol...i m just so dumb...

Originally Posted by Calculus26
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

9. 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

Originally Posted by Calculus26
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

10. ## using logarithm..to solve y..need help!! thx

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

11. Originally Posted by mookie
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
$-e^{-y}=-\frac{1}{2}e^{-x^2}+c$

$e^{-y}=\frac{1}{2}e^{-x^2}+c$

$\ln(e^{y})=\ln(\frac{1}{2}e^{-x^2}+c)$

$y=\ln(\frac{1}{2}e^{-x^2}+c)$

Now using the IC

$0=\ln(\frac{1}{2}e^{0}+c) \iff 0=\ln(\frac{1}{2}+c) \iff 1=\frac{1}{2}+c \iff \frac{1}{2}=c$