dx/dy cos x sin y + x^5 = 11
i got dx/dy= (-cos x cos y)/ (-sin x sin y +5x^4)
But the wolfram got
(1- cos x cos y)/(5x^4-sin y sin x), where is the 1 come from?
so we are differentiating
cosx siny + x^5 = 11 implicitly ?
cosx(cosy) dy/dx + siny(-sinx) + 5x^4 = 0
dy/dx = (sinx siny - 5x^3)/(cosx cosy)
which agrees with my Wolfram result
http://www.wolframalpha.com/input/?i=find+dy%2Fdx+for+%28cos%28x%29%29%28sin%28y%29%29+%2B+x%5E5+%3D+11
just noticed you wanted dx/dy (strange)
cosx siny + x^5 = 11
cosx(cosy) + siny(-sinx) dx/dy = 5x^4 dx/dy = 0
dx/dy (5x4 - sinxsiny) = - cosxcosy
dx/dy = -cosxcosy/(5x^4 - sinxsiny)
the same as yours.
At reiny .... Thanks
To understand where the "1" comes from in the Wolfram Alpha solution, let's go through the steps of solving the differential equation.
First, we have the equation:
dx/dy * cos(x) * sin(y) + x^5 = 11
To solve this, we can consider dx/dy as the independent variable, denoted by u. So, our equation becomes:
u * cos(x) * sin(y) + x^5 = 11
Now, let's differentiate both sides of the equation with respect to y:
du/dy * cos(x) * sin(y) + u * cos(x) * cos(y) + 5x^4 * dx/dy = 0
Next, let's isolate du/dy by subtracting u * cos(x) * cos(y) from both sides:
du/dy * cos(x) * sin(y) + 5x^4 * dx/dy = -u * cos(x) * cos(y)
Now, we can rearrange terms to group dx/dy and du/dy separately:
du/dy * cos(x) * sin(y) = -5x^4 * dx/dy - u * cos(x) * cos(y)
Dividing both sides by cos(x) * sin(y), we get:
du/dy = (-5x^4 * dx/dy - u * cos(x) * cos(y)) / (cos(x) * sin(y))
Now, recall that we initially denoted dx/dy as u. So we have:
du/dy = (-5x^4 * u - u * cos(x) * cos(y)) / (cos(x) * sin(y))
Factoring out the common term of u on the numerator:
du/dy = u * (-5x^4 - cos(x) * cos(y)) / (cos(x) * sin(y))
Finally, we can divide both sides by (-5x^4 - cos(x) * cos(y)) to solve for u, which gives us:
u = 1 / (5x^4 + cos(x) * cos(y))
Therefore, the solution for dx/dy is:
dx/dy = u = 1 / (5x^4 + cos(x) * cos(y))
So, the "1" comes from isolating du/dy and dividing both sides of the equation by (-5x^4 - cos(x) * cos(y)).