# Answer to Question #23827 in Differential Equations for Matthew Lind

Question #23827

Show that there exists a unique solution for the system:

U_x = 3x^2y+y

U_y = x^3+x

together with the initial condition u(0,0) = 0.

U_x = 3x^2y+y

U_y = x^3+x

together with the initial condition u(0,0) = 0.

Expert's answer

U_x = 3x^2y+y

U_y = x^3+x

Integrate the first:

U = x^3 *y + y*x + f(y) where f(y) - some function.

Substitute it to the second equation:

(x^3 *y + y*x + f(y))_y = x^3+x

x^3 + x + f '(y) = x^3 + x

Therefore f '(y) = 0 => f(y) = const.

The initial condition U(0,0) = 0.

U = x^3 *y + y*x + const

U (0, 0) = const = 0

U (x, y) = x^3 *y + y*x

Answer: U (x, y) = x^3 *y + y*x.

U_y = x^3+x

Integrate the first:

U = x^3 *y + y*x + f(y) where f(y) - some function.

Substitute it to the second equation:

(x^3 *y + y*x + f(y))_y = x^3+x

x^3 + x + f '(y) = x^3 + x

Therefore f '(y) = 0 => f(y) = const.

The initial condition U(0,0) = 0.

U = x^3 *y + y*x + const

U (0, 0) = const = 0

U (x, y) = x^3 *y + y*x

Answer: U (x, y) = x^3 *y + y*x.

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

## Leave a comment