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Solve x(x-y)dy+y^2dx=0 using

User CFinck
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Answer:

the method of exact differential equations:

We need to check if this differential equation is exact. To do that, we check if the partial derivative of the first term with respect to y is equal to the partial derivative of the second term with respect to x:

∂/∂y(x(x-y)) = x(-1) = -x

∂/∂x(y^2) = 0

Since these partial derivatives are not equal, the differential equation is not exact. We can try to make it exact by multiplying the entire equation by a suitable integrating factor.

Let's find the integrating factor (IF) by taking the partial derivative of the IF with respect to y and equating it to the partial derivative of the second term with respect to x:

∂/∂y(IF) = -y^2/(x(x-y)^2)

∂/∂x(y^2) = 0

From the first equation, we can see that an integrating factor of IF = x(x-y)^2 should make the equation exact. Multiplying the entire equation by this integrating factor, we get:

x(x-y)^2dy + y^2x(x-y)dx = 0

Now, we just need to find a function φ(x,y) such that:

∂φ/∂x = x(x-y)^2dy

∂φ/∂y = y^2x(x-y)dx

Integrating the first equation with respect to x, we get:

φ(x,y) = ∫x(x-y)^2dy + f(x)

φ(x,y) = -1/3(x-y)^3x + f(x)

Now, we differentiate this equation with respect to x and equate it to the second equation:

∂φ/∂x = -(x-y)^3 + f'(x)

∂φ/∂y = y^2(x-y)^3

Comparing the two, we can see that f'(x) = 0, which means that f(x) is a constant. We can choose this constant to be zero without loss of generality.

Therefore, the solution to the differential equation is given by:

-1/3(x-y)^3x + C = 0

where C is an arbitrary constant.

User Flowryn
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