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Popular >

y^{''}-4y^'+5=0

Solution

y^{^{''}} - 4 y^{^{'}} + 5 = 0

Solution

y = c_{1} e^{4 t} + c_{2} + \frac{5 t}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} + \frac{5 t}{4}

y = c_{1} e^{4 t} + c_{2} + \frac{5 t}{4}

Popular Examples

tangent of f(x)= 5/(2sqrt(5x+6)),\at a=2 tangent of

f (x) = \frac{5}{2 5 x + 6}, at a = 2

area y=(x^3)/9 ,0<= x<= 2 area

y = \frac{x ^{3}}{9}, 0 \leq x \leq 2

ty^'+7y=t^2-t+1,y(1)= 1/7

t y^{^{'}} + 7 y = t^{2} - t + 1, y (1) = \frac{1}{7}

integral of tan^2(x

\int tan^{2} (d) x d x

derivative of ln(x)

\frac{d}{d x} (ln (x))

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-4y^'+5=0 ?
The solution for y^{''}-4y^'+5=0 is y=c_{1}e^{4t}+c_{2}+(5t)/4