What is the solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=5 ?

The solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=5 is y=(1+7/4)e^{(3t)/2}-7/4 e^{-t/2}

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4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=5

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y = (1 + \frac{7}{4}) e^{\frac{3 t}{2}} - \frac{7}{4} e^{- \frac{t}{2}}

Solution steps

4 y^{''} (t) - 4 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = (1 + \frac{7}{4}) e^{\frac{3 t}{2}} - \frac{7}{4} e^{- \frac{t}{2}}

y = (1 + \frac{7}{4}) e^{\frac{3 t}{2}} - \frac{7}{4} e^{- \frac{t}{2}}

x \to 0 + lim ((cot (x))^{s i n (x)})

\frac{d y}{d t} + 2 (t + 1) y^{2} = 0

\int 15 sin^{3} (x) cos^{2} (x) d x

x^{2} \frac{d y}{d x} - 2 x y = 3 y^{4}

\int \frac{5}{x} d x

What is the solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=5 ?
The solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=5 is y=(1+7/4)e^{(3t)/2}-7/4 e^{-t/2}