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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 5 y^{^{'}} - 2 a^{2} y = 0

x^{^{''}} + 2 \cdot b \cdot x^{^{'}} + x = 0, x (0) = 1, x^{^{'}} (0) = - 1

y^{^{''}} - 4 π^{2} y = 0, y (0) = 0, y^{^{'}} (1) = 0

\frac{d ^{2} x}{d t ^{2}} = a (x)

x^{^{''}} - 3 x^{^{'}} - 3 x = 0, x (0) = 1

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