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

The solution for y^{''}-3y^'+2y=a,y(0)=2,y^'(0)=-6 is y=(2-(-a+10)-a/2)e^{2t}+(-a+10)e^t+a/2

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

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

y = (2 - (- a + 10) - \frac{a}{2}) e^{2 t} + (- a + 10) e^{t} + \frac{a}{2}

Solution steps

y^{''} (t) - 3 y^{'} (t) + 2 y = a

Solve linear ODE:

y = (2 - (- a + 10) - \frac{a}{2}) e^{2 t} + (- a + 10) e^{t} + \frac{a}{2}

y = (2 - (- a + 10) - \frac{a}{2}) e^{2 t} + (- a + 10) e^{t} + \frac{a}{2}

d (x) = \frac{2500}{30 ( 0.3 + 0.03 )}

y^{^{''}} = 2 y - x

y^{^{''}} - y \cdot a + g = 0

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 36 e^{x}, y^{^{'}} (0) = 26

y^{^{''}} - 2 y^{^{'}} + 2 y = e x sec (x)

What is the solution for y^{''}-3y^'+2y=a,y(0)=2,y^'(0)=-6 ?
The solution for y^{''}-3y^'+2y=a,y(0)=2,y^'(0)=-6 is y=(2-(-a+10)-a/2)e^{2t}+(-a+10)e^t+a/2