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AP Calculus Problem! Help!

"Find the General Solution for each differential equation. "

It's multivariable, so it can't be integrated normally.



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


please explain how, so I can finish the rest of my homework

User Tajihiro
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2 Answers

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Final answer:

The general solution to the differential equation dy/dx = x^2/y^2 is found by separating variables, integrating both sides, and simplifying, resulting in the solution y^3 = x^3 + k, where k is an arbitrary constant.

Step-by-step explanation:

To find the general solution for the given differential equation dy/dx = x2/y2, we can use separation of variables. We'll take all the y terms to one side of the equation and all the x terms to the other side.

First, we rewrite the equation as y2 dy = x2 dx. Now, we integrate both sides to find the antiderivatives. The integral of y2 with respect to y is (1/3)y3, and the integral of x2 with respect to x is (1/3)x3. Remembering to add the constant of integration, we have:

(1/3)y3 = (1/3)x3 + C

Multiplying through by 3 to clear the fractions gives us:

y3 = x3 + 3C

The constant 3C can be simplified to a new constant, which we'll call k. So, the general solution to the differential equation is:

y3 = x3 + k

Note: The constant k represents an arbitrary constant that can be determined if an initial condition is given.

User Mperle
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The equation is separable. That is to say, we're given a differential equation of the form


(\mathrm dy)/(\mathrm dx)=f(x,y)

but it happens that we can write
f(x,y)=g(x)h(y), a product of functions of their own independent variables. When this is the case, we can split up the differentials:


(\mathrm dy)/(\mathrm dx)=g(x)h(y)\implies(\mathrm dy)/(h(y))=g(x)\,\mathrm dx

Then we integrate both sides and attempt to solve for
y(x).

For this ODE, we can write


(\mathrm dy)/(\mathrm dx)=(x^2)/(y^2)\implies y^2\,\mathrm dy=x^2\,\mathrm dx


\implies\displaystyle\int y^2\,\mathrm dy=\int x^2\,\mathrm dx


\implies\frac{y^3}3=\frac{x^3}3+C


\implies y^3=x^3+C


\implies y=(x^3+C)^(1/3)=\sqrt[3]{x^3+C}

Other methods are available, but I think they tend to be outside the scope of AP Calc curriculum.

User Reub
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7.5k points
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