What is the solution for y^{''}+y=x^2+1,y(0)=4,y(1)=0 ?

The solution for y^{''}+y=x^2+1,y(0)=4,y(1)=0 is y=5cos(x)-(5cos(1)sin(x))/(sin(1))+x^2-1

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

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

y = 5 cos (x) - \frac{5 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

Solution steps

y^{''} (x) + y = x^{2} + 1

Solve linear ODE:

y = 5 cos (x) - \frac{5 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

y = 5 cos (x) - \frac{5 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

y^{^{''}} + y = tan (x), y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} + y^{^{'}} = - 2 cos (t)

x^{^{''}} - 5 x^{^{'}} + 6 x = t^{2} e^{3 t}

\frac{d ^{2} y}{d x ^{2}} - 9 y = sin (3 x)

y^{^{''}} + 16 y = cos (4 t), y (0) = 6, y^{^{'}} (0) = 8

What is the solution for y^{''}+y=x^2+1,y(0)=4,y(1)=0 ?
The solution for y^{''}+y=x^2+1,y(0)=4,y(1)=0 is y=5cos(x)-(5cos(1)sin(x))/(sin(1))+x^2-1