Answer:
[tex]m + b=2[/tex]
Step-by-step explanation:
From the question we are told that
The non separable differential equation [tex]\frac{dy}{dx} = 8x - 2y[/tex]
Generally the chi form of a give equation is mathematically represented as
[tex]\frac{dy}{dx} +P(x)y= q(x)[/tex]
[tex]\frac{dy}{dx} + 2y = 8x[/tex]
[tex]ye^2^x=\int e^2^.8xdxx[/tex]
[tex]ye^2^x=8(\frac{xe^2^x}{2}-\frac{\int e^2^x}{2})[/tex]
[tex]ye^2^x=8(\frac{xe^2^x}{2}-\frac{e^2^x}{1})+c\\[/tex]
[tex]ye^2^x=4x^2^x-2e^2^x+c[/tex]
[tex]y=4x-2+c[/tex]
for c=0
[tex]y=4x-2[/tex]
Therefore
[tex]m=4\\b=-2\\m + b=4-(+2)[/tex]
[tex]m + b=2[/tex]
The linear particular solution of the form [tex]y=mx+b[/tex] where [tex]m+b=4-2=2[/tex]
Nonseparable differential equation:
Differential equations that cannot be written [tex]\frac{dy}{dx} =f(x)g(y)[/tex] are non-separable differential equations.
In other words, we can say those differential equations which are not separable are called non-separable differential equations.
The given equation is,
[tex]\frac{dy}{dx}=8x-2y[/tex]
This is in the form [tex]\frac{dy}{dx} +p(x)y=\phi(x)[/tex]
Here [tex]P=2, Q=8x[/tex]
So, [tex]e^{\int_{}^{}Pdx}=e^{\int_{}^{}2dx}=e^{2x}[/tex]
Therefore, the general solution is,
[tex]ye^{\int_{}^{}Pdx} =\int_{}^{}Qe^{\int_{}^{}Pdx}dx+c\\ye^{2x}=\int_{}^{}8xe^{2x}dx+c\\y=4x-2+c^{-2x}[/tex]
Learn more about the topic of nonseparable differential equations:
https://brainly.com/question/25731911
