Solve the differential equation dy/dx = -xe^y and determine the equation of the curve through P(1,2)

I tried solving the differential equation and I get y = log(x^2/2 + C).
Is this correct?

Now I forgot how to find the equation.

Thank you!

Your solution is almost good, just a change of the sign will fix it.

y=-log(x^2/2+C)

If it has to pass through P(1,2)
substitute x=1, and y=2 and find C.
2=-log(1/2+C)
log(1/2+C)=-2
take logs
1/2+C=e^(-2)
C=e^(-2)-1/2

so
y=-log(x²/2+e^(-2)-1/2)

thanks a lot mate

oops, that was under my brother's name hehe

You're both welcome!

To solve the given differential equation dy/dx = -xe^y, we can separate the variables and integrate both sides with respect to their respective variables.

Step 1: Start with the given differential equation dy/dx = -xe^y.

Step 2: Separate the variables by multiplying both sides by dx and dividing by e^y:

dy / e^y = -x dx.

Step 3: Integrate both sides with respect to their respective variables:

∫ (dy / e^y) = ∫ (-x dx).

Step 4: Evaluate the integrals:

-ln|e^y| = -x^2/2 + C.

Simplifying this expression, we get:

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

Step 5: Rewrite the natural logarithm of the exponential:

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

Step 6: Simplify further:

y = -x^2/2 + C.

Now, let's find the equation of the curve passing through the point P(1, 2).

To find the value of C, substitute the coordinates of the point P(1, 2) into the equation:

2 = -(1^2)/2 + C.

Simplifying this equation, we get:

2 = -1/2 + C.

Adding 1/2 to both sides, we have:

2 + 1/2 = C.

Simplifying further, we find:

C = 5/2.

Therefore, the equation of the curve passing through P(1, 2) is:

y = -x^2/2 + 5/2.