What is the solution for y^{''}-10y^'+25=0 ?

The solution for y^{''}-10y^'+25=0 is y=c_{1}e^{10t}+c_{2}+(5t)/2

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y^{''}-10y^'+25=0

y^{^{''}} - 10 y^{^{'}} + 25 = 0

y = c_{1} e^{10 t} + c_{2} + \frac{5 t}{2}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 = 0

Solve linear ODE:

y = c_{1} e^{10 t} + c_{2} + \frac{5 t}{2}

y = c_{1} e^{10 t} + c_{2} + \frac{5 t}{2}

\frac{d}{d x} ((\frac{cos ( x )}{1 + sin ( x )})^{5})

\int \frac{x ^{3} + 2 x ^{2} + 3 x - 2}{x - 2} d x

x^{\frac{11}{2}}, 1, x = 1.4

\int \frac{2}{x} - \frac{4}{x ^{3}} d x

3 sin^{2} (t)

What is the solution for y^{''}-10y^'+25=0 ?
The solution for y^{''}-10y^'+25=0 is y=c_{1}e^{10t}+c_{2}+(5t)/2