What is the solution for y^{''}+y^'=0,y(pi/4)=0,y^'(pi/4)=-3 ?

The solution for y^{''}+y^'=0,y(pi/4)=0,y^'(pi/4)=-3 is y=-3+3e^{pi/4-t}

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y^{''}+y^'=0,y(pi/4)=0,y^'(pi/4)=-3

y^{^{''}} + y^{^{'}} = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3

y = - 3 + 3 e^{\frac{π}{4} - t}

Solution steps

y^{''} (t) + y^{'} (t) = 0

Solve linear ODE:

y = - 3 + 3 e^{\frac{π}{4} - t}

y = - 3 + 3 e^{\frac{π}{4} - t}

y^{^{''}} + (\frac{v}{a}) y^{^{'}} = 0

\frac{d ^{4} y}{d x ^{4}} - 2 (\frac{d ^{2} y}{d x ^{2}}) + y = 0

y^{^{''}} - 16 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} - 7 y^{^{'}} = 0, y (0) = 2 - 2 e^{7}, y (1) = 0

\frac{d ^{2} y}{d x ^{2}} + 15 \frac{d y}{d x} + 56 y = 0

What is the solution for y^{''}+y^'=0,y(pi/4)=0,y^'(pi/4)=-3 ?
The solution for y^{''}+y^'=0,y(pi/4)=0,y^'(pi/4)=-3 is y=-3+3e^{pi/4-t}