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.