What is the solution for y^{''}+y=t,y(0)=1,y(1)=-2 ?

The solution for y^{''}+y=t,y(0)=1,y(1)=-2 is y=cos(t)+((-cos(1)-3)sin(t))/(sin(1))+t

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y^{''}+y=t,y(0)=1,y(1)=-2

y^{^{''}} + y = t, y (0) = 1, y (1) = - 2

y = cos (t) + \frac{( - cos ( 1 ) - 3 ) sin ( t )}{sin ( 1 )} + t

Solution steps

y^{''} (t) + y = t

Solve linear ODE:

y = cos (t) + \frac{( - cos ( 1 ) - 3 ) sin ( t )}{sin ( 1 )} + t

y = cos (t) + \frac{( - cos ( 1 ) - 3 ) sin ( t )}{sin ( 1 )} + t

h \to 0 lim (\frac{( 3 + h ) ^{4} - 81}{h})

\frac{d}{d x} (\frac{1}{2 x} + \frac{1}{8 x ^{\frac{7}{8}}})

e^{- 2 t} (2 cos (3 t) - sin (3 t))

x \to \infty lim (\frac{5}{x})

\int_{1}^{4} \frac{x + 4}{2 x} d x

What is the solution for y^{''}+y=t,y(0)=1,y(1)=-2 ?
The solution for y^{''}+y=t,y(0)=1,y(1)=-2 is y=cos(t)+((-cos(1)-3)sin(t))/(sin(1))+t