Question

Solve the initial value problem yy' + x = √x² + y² with y(5) = -√56.

To solve this, we should use the substitution
u = x²+y² help (formulas)
u' = 2x+2yy' help (formulas)

Enter derivatives using prime notation (e.g., you would enter y' for dy/dx)

Answer 1

The solution to the initial value problem
`\(yy' + x = √(x^2 + y^2)\)` with

`\(y(5) = -√(56)\)` is
`\(y = \pm √(x^2 + C)\)`, where C = 31.

Let's simplify the given differential equation using the provided substitution
`\(u = x^2 + y^2\)`:

1. Find the derivatives u' and y':

`\[u' = 2x + 2yy'\]`

`\[y' = (u' - 2x)/(2y)\]`

2. Substitute into the given differential equation:

`\[y\left((u' - 2x)/(2y)\right) + x = √(x^2 + y^2)\]`

3. Simplify by canceling common terms:

`\[(1)/(2)u' - x = √(x^2 + y^2)\]`

4. Square both sides to eliminate the square root:

`\[(1)/(4)u'^2 - xu' + x^2 = x^2 + y^2\]`

5. Substitute back
`\(u = x^2 + y^2\)`:

`\[(1)/(4)u'^2 - xu' + u = u\]`

6. Simplify further:

`\[(1)/(4)u'^2 - xu' = 0\]`

7. Solve for u':

u' - 4x = 0

u' = 4x

8. Integrate u' with respect to x to find u:

`\[u = 2x^2 + C\]`

9. Substitute back
`\(u = x^2 + y^2\):`

`\[x^2 + y^2 = 2x^2 + C\]`

10. Solve for y:

`\[y^2 = x^2 + C\]`

`\[y = \pm √(x^2 + C)\]`

This is the solution to the given initial value problem, where C is the constant of integration.