What is the solution for y^{''}-20y^'+100y=0,y(0)=4,y(1)=3 ?

The solution for y^{''}-20y^'+100y=0,y(0)=4,y(1)=3 is y=4e^{10t}+(3-4e^{10})e^{10t-10}t

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y^{''}-20y^'+100y=0,y(0)=4,y(1)=3

y^{^{''}} - 20 y^{^{'}} + 100 y = 0, y (0) = 4, y (1) = 3

y = 4 e^{10 t} + (3 - 4 e^{10}) e^{10 t - 10} t

Solution steps

y^{''} (t) - 20 y^{'} (t) + 100 y = 0

Solve linear ODE:

y = 4 e^{10 t} + (3 - 4 e^{10}) e^{10 t - 10} t

y = 4 e^{10 t} + (3 - 4 e^{10}) e^{10 t - 10} t

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 10 y = 0

y^{^{''}} + 8 y^{^{'}} - 2 y = 0

y^{^{''}} + 4 y = 0, y (0) = y^{^{'}} (0)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 3 y = 0

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d ^{2} y}{d x ^{2}} + 8 y = 0

What is the solution for y^{''}-20y^'+100y=0,y(0)=4,y(1)=3 ?
The solution for y^{''}-20y^'+100y=0,y(0)=4,y(1)=3 is y=4e^{10t}+(3-4e^{10})e^{10t-10}t