Question

What's the particular solution that satisfies `y(0)=0` of `dy/dx=(-2x+y)^2-7` ?

Answer 1

`y = (2z+3 -2xe^(6x) + 3e^(6x))/(1-e^(6x))`

Explanation:

We have:

`dy/dx=(-2x+y)^2-7` with `y(0)=0`

We can perform a substitution:

Let`u=-2x+y \` and `\ x=0,y=0 => u=0`

If we differentiate wrt `x`, then:

`(du)/dx = -2+dy/dx => dy/dx = (du)/dx +2`

Substituting into the initial DE:

`(du)/(dx) + 2 = u^2-7` with `u(0)=0`

`:. (du)/(dx) = u^2-9`

`:. 1/(u^2-9) (du)/(dx) = 1`

This has reduced the equation to a First Order Separable ODE, so we can "separate the variable":

`int \ 1/(u^2-9) \ du =int \ u \ du`

To integrate the LHS we can decompose the integrand into partial fractions:

`1/(u^2-9) -= 1/((u-3)(u+3)) -= A/(u-3) + B/(u+3)`

Using the "cover-up" method we then have:

`1/((u-3)(u+3)) -= (1/6)/(u-3) + (-1/6)/(u+3)`

So then we have:

`1/6 int \ 1/(u-3) - 1/(u+3) \ du = int \ dx`

Which we can now readily integrate:

`1/6 {ln|u-3| - ln |u+3|} = x + C`

Applying the initial condition `u(0)=0`

`1/6 {ln|-3| - ln |3|} = C => C = 0`

So we get a Particular Solution:

`ln|u-3| - ln |u+3| = 6x`

`:. ln|(u-3)/(u+3)| = 6x`

`:. |(u-3)/(u+3)| = e^(6x)`

Noting that `e^a gt 0 AA a in RR`

`(u-3)/(u+3) = e^(6x)`

`:. u-3 = ue^(6x) +3 e^(6x)`

Restoring the earlier substitution `u=-2x+y`

`-2x+y-3 = (-2x+y)e^(6x) +3e^(6x)`

`:. -2x+y-3 = -2xe^(6x) +ye^(6x) + 3e^(6x)`

`:. y - ye^(6x) = 2z+3 -2xe^(6x) + 3e^(6x)`

`:. y = (2z+3 -2xe^(6x) + 3e^(6x))/(1-e^(6x))`