What is the solution for y^{''}+2x=6x^2,y(0)=3 ?

The solution for y^{''}+2x=6x^2,y(0)=3 is y=(x^4)/2-(x^3)/3+c_{1}x+3

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y^{''}+2x=6x^2,y(0)=3

y^{^{''}} + 2 x = 6 x^{2}, y (0) = 3

y = \frac{x ^{4}}{2} - \frac{x ^{3}}{3} + c_{1} x + 3

Solution steps

y^{''} (x) + 2 x = 6 x^{2}

Isolate

y^{''} (x) : y^{''} (x) = 6 x^{2} - 2 x

Solve

y^{''} (x) = 6 x^{2} - 2 x : y^{'} (x) = 2 x^{3} - x^{2} + c_{1}

Solve

y^{'} (x) = 2 x^{3} - x^{2} + c_{1} : y = \frac{x ^{4}}{2} - \frac{x ^{3}}{3} + c_{1} x + c_{2}

Apply initial conditions:

y = \frac{x ^{4}}{2} - \frac{x ^{3}}{3} + c_{1} x + 3

y = \frac{x ^{4}}{2} - \frac{x ^{3}}{3} + c_{1} x + 3

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What is the solution for y^{''}+2x=6x^2,y(0)=3 ?
The solution for y^{''}+2x=6x^2,y(0)=3 is y=(x^4)/2-(x^3)/3+c_{1}x+3