Substitute [tex]v(x)=-2x+y(x)[/tex], so that [tex]\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}[/tex]. Then the ODE is equivalent to
[tex]\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7[/tex]
which is separable as
[tex]\dfrac{\mathrm dv}{v^2-9}=\mathrm dx[/tex]
Split the left side into partial fractions,
[tex]\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)[/tex]
so that integrating both sides is trivial and we get
[tex]\dfrac{\ln|v-3|-\ln|v+3|}6=x+C[/tex]
[tex]\ln\left|\dfrac{v-3}{v+3}\right|=6x+C[/tex]
[tex]\dfrac{v-3}{v+3}=Ce^{6x}[/tex]
[tex]\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}[/tex]
[tex]\dfrac6{v+3}=1-Ce^{6x}[/tex]
[tex]v=\dfrac6{1-Ce^{6x}}-3[/tex]
[tex]-2x+y=\dfrac6{1-Ce^{6x}}-3[/tex]
[tex]y=2x+\dfrac6{1-Ce^{6x}}-3[/tex]
Given the initial condition [tex]y(0)=0[/tex], we find
[tex]0=\dfrac6{1-C}-3\implies C=-1[/tex]
so that the ODE has the particular solution,
[tex]\boxed{y=2x+\dfrac6{1+e^{6x}}-3}[/tex]
