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)})

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

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

\frac{d}{d x} (e^{(- x^{\frac{1}{2}} + \frac{1}{5} x^{- \frac{3}{5}})})

\int \frac{1}{( x ^{2} + 100 ) ^{2}} 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}