Question

Y' = 1 + x ^ 2 + y ^ 2 + x ^ 2 * y ^ 2 , y(0) = 1

Answer 1

y = tan(x³/3 +x +π/4)

Explanation:

You want the solution to the differential equation ...

y' = 1 + x^2 + y^2 + x^2*y^2 with y(0) = 1.

Separation of variables

We can rewrite the equation and separate the variables for integration.

`y'=(1+x^2)(1+y^2)\\\\(dy)/(1+y^2)=(1+x^2)dx`

Integration

Integrating both sides of this equation gives ...

arctan(y) = x +x³/3 +C

y = tan(x +x³/3 +C)

Initial condition

To satisfy the initial condition y(0) = 1, we must have ...

1 = tan(0 +C)

C = arctan(1) = π/4

Then the solution to the differential equation is ...

y = tan(x³/3 +x +π/4)

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