What is the solution for y^'-2x=4x^3,y(0)=1 ?

The solution for y^'-2x=4x^3,y(0)=1 is y=x^4+x^2+1

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y^'-2x=4x^3,y(0)=1

y^{^{'}} - 2 x = 4 x^{3}, y (0) = 1

y = x^{4} + x^{2} + 1

Solution steps

y^{'} (x) - 2 x = 4 x^{3}

Isolate

y^{'} (x) : y^{'} (x) = 4 x^{3} + 2 x

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 4 x^{3} + 2 x d x

\int 4 x^{3} + 2 x d x = x^{4} + x^{2} + c_{1}

y = x^{4} + x^{2} + c_{1}

Apply initial conditions:

y = x^{4} + x^{2} + 1

y = x^{4} + x^{2} + 1

What is the solution for y^'-2x=4x^3,y(0)=1 ?
The solution for y^'-2x=4x^3,y(0)=1 is y=x^4+x^2+1