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Solve the initial value problem xy²(dy/dx)=y³-x³ , with the initial condition y(1)=4. What is the solution?

A) y=x²
B) y=1/x²
C) y=√x
D) y=e^x

1 Answer

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Final answer:

To solve the initial value problem xy²(dy/dx)=y³-x³, we need to separate the variables and integrate both sides of the equation using the given initial condition. The solution to the initial value problem is y = (x^3 + (x/√63)^2)^(1/3).

Step-by-step explanation:

To solve the initial value problem, we need to separate the variables and integrate both sides of the equation. Starting with the given equation: xy^2(dy/dx) = y^3 - x^3. We can rewrite it as: 1/(y^3 - x^3) * y^2 dy = 1/x dx.

Integrating both sides, we get: ∫1/(y^3 - x^3) * y^2 dy = ∫1/x dx. Simplifying, we have:

-1/2 * ln|y^3 - x^3| = ln|x| + C, where C is the constant of integration.

Applying the initial condition y(1) = 4, we can determine the value of C:

-1/2 * ln|4^3 - 1^3| = ln|1| + C,

-1/2 * ln|63| = 0 + C,

C = -1/2 * ln|63|.

Substituting this value into the equation, we have:

-1/2 * ln|y^3 - x^3| = ln|x| - 1/2 * ln|63|.

Taking the exponential of both sides, we get:

e^(-1/2 * ln|y^3 - x^3|) = e^(ln|x| - 1/2 * ln|63|).

Simplifying, we have:

|y^3 - x^3|^(-1/2) = |x|/√63.

Squaring both sides, we get:

|y^3 - x^3|^(-1) = (|x|/√63)^2,

which simplifies to:

|y^3 - x^3| = (√63/|x|)^2.

Since the absolute value can be eliminated by taking both positive and negative solutions, we have:

y^3 - x^3 = (|x|/√63)^2 or y^3 - x^3 = -(|x|/√63)^2.

For simplicity, let's consider only the positive solution: y^3 - x^3 = (x/√63)^2.

Rearranging the equation, we have:

y^3 = x^3 + (x/√63)^2.

Taking the cube root of both sides, we get:

y = (x^3 + (x/√63)^2)^(1/3).

Therefore, the solution to the initial value problem is: y = (x^3 + (x/√63)^2)^(1/3).

User Jamezor
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