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

The solution for y^{''}-2y^'+3=e^t,y(0)=1,y^'(0)=2 is y=(-5/4+2)e^{2t}+5/4-e^t+(3t)/2

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y^{''}-2y^'+3=e^t,y(0)=1,y^'(0)=2

y^{^{''}} - 2 y^{^{'}} + 3 = e^{t}, y (0) = 1, y^{^{'}} (0) = 2

y = (- \frac{5}{4} + 2) e^{2 t} + \frac{5}{4} - e^{t} + \frac{3 t}{2}

Solution steps

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

Solve linear ODE:

y = (- \frac{5}{4} + 2) e^{2 t} + \frac{5}{4} - e^{t} + \frac{3 t}{2}

y = (- \frac{5}{4} + 2) e^{2 t} + \frac{5}{4} - e^{t} + \frac{3 t}{2}

2 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 = 0

(d^{3} - 3 d^{2} + 4 d - 2) y = x^{3}

y^{^{''}} = 2 y + tan^{3} (x), y (0) = \frac{1}{2} tan (x)

x^{^{''}} + 9 x = 10 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 10 y = 30 t + 9, y (0) = 7, y^{^{'}} (0) = 26

What is the solution for y^{''}-2y^'+3=e^t,y(0)=1,y^'(0)=2 ?
The solution for y^{''}-2y^'+3=e^t,y(0)=1,y^'(0)=2 is y=(-5/4+2)e^{2t}+5/4-e^t+(3t)/2