What is the solution for y^{''}-5y^'+4y=0,y(0)=6,y(1)=3 ?

The solution for y^{''}-5y^'+4y=0,y(0)=6,y(1)=3 is y=(6-(6e^4-3)/(e^4-e))e^{4t}+(6e^4-3)/(e^4-e)e^t

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y^{''}-5y^'+4y=0,y(0)=6,y(1)=3

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 6, y (1) = 3

y = (6 - \frac{6 e ^{4} - 3}{e ^{4} - e}) e^{4 t} + \frac{6 e ^{4} - 3}{e ^{4} - e} e^{t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = (6 - \frac{6 e ^{4} - 3}{e ^{4} - e}) e^{4 t} + \frac{6 e ^{4} - 3}{e ^{4} - e} e^{t}

y = (6 - \frac{6 e ^{4} - 3}{e ^{4} - e}) e^{4 t} + \frac{6 e ^{4} - 3}{e ^{4} - e} e^{t}

n = 0 \sum \infty \frac{2 ^{n}}{n} x^{n}

\int sin (2 x) sec (x) d x

y = x^{10 c o s (x)}

y = 4 tan (x), at x = \frac{π}{6}

\int_{1}^{3} (x^{2} + 1)^{3} x d x

What is the solution for y^{''}-5y^'+4y=0,y(0)=6,y(1)=3 ?
The solution for y^{''}-5y^'+4y=0,y(0)=6,y(1)=3 is y=(6-(6e^4-3)/(e^4-e))e^{4t}+(6e^4-3)/(e^4-e)e^t