Question

Find the solution of the initial value problem

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

Answer 1

Substitute
`v(x)=-2x+y(x)`, so that
`(\mathrm dv)/(\mathrm dx)=-2+(\mathrm dy)/(\mathrm dx)`. Then the ODE is equivalent to

`(\mathrm dv)/(\mathrm dx)+2=v^2-7`

which is separable as

`(\mathrm dv)/(v^2-9)=\mathrm dx`

Split the left side into partial fractions,

`\frac1{v^2-9}=\frac16\left(\frac1{v-3}-\frac1{v+3}\right)`

so that integrating both sides is trivial and we get

`\frac\ln6=x+C`

`\ln\left|(v-3)/(v+3)\right|=6x+C`

`(v-3)/(v+3)=Ce^(6x)`

`(v+3-6)/(v+3)=1-\frac6{v+3}=Ce^(6x)`

`\frac6{v+3}=1-Ce^(6x)`

`v=\frac6{1-Ce^(6x)}-3`

`-2x+y=\frac6{1-Ce^(6x)}-3`

`y=2x+\frac6{1-Ce^(6x)}-3`

Given the initial condition
`y(0)=0`, we find

`0=\frac6{1-C}-3\implies C=-1`

so that the ODE has the particular solution,

`\boxed{y=2x+\frac6{1+e^(6x)}-3}`