What is the solution for y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)=-3/4 ?

The solution for y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)=-3/4 is y=(1/2-7/4)e^{2t}+7/4 e^t

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y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)=-3/4

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = - \frac{3}{4}

y = (\frac{1}{2} - \frac{7}{4}) e^{2 t} + \frac{7}{4} e^{t}

Solution steps

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

Solve linear ODE:

y = (\frac{1}{2} - \frac{7}{4}) e^{2 t} + \frac{7}{4} e^{t}

y = (\frac{1}{2} - \frac{7}{4}) e^{2 t} + \frac{7}{4} e^{t}

x \to 0 lim (\frac{( \frac{1}{2} ) ^{x} - 1}{x})

f (x) = sec (2 x)

\int e^{x + e^{x}} d x

\frac{1}{s ^{2}} - \frac{48}{s ^{5}}

\int \frac{1}{x x ^{2} - 2} d x

What is the solution for y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)=-3/4 ?
The solution for y^{''}-3y^'+2y=0,y(0)= 1/2 ,y^'(0)=-3/4 is y=(1/2-7/4)e^{2t}+7/4 e^t