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Popular >

y^{''}-ay^'=0

Solution

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

Solution

y = c_{1} e^{a t} + c_{2}

Solution steps

y^{''} (t) - a y^{'} (t) = 0

Solve linear ODE:

y = c_{1} e^{a t} + c_{2}

y = c_{1} e^{a t} + c_{2}

Popular Examples

tangent of 5=(x+9)^2+(y-2)^2,(-7,3)tangent of

5 = (x + 9)^{2} + (y - 2)^{2}, (- 7, 3)

e^{2x}(dy)/(dx)-e^xy=e^{-x}y^2

e^{2 x} \frac{d y}{d x} - e^{x} y = e^{- x} y^{2}

derivative of (-x/(sqrt(1+x^2)))

\frac{d}{d x} (\frac{- x}{1 + x ^{2}})

area y=x^3-x^2,y=0 area

y = x^{3} - x^{2}, y = 0

(\partial)/(\partial x)(3x^2y^4+1)

\frac{\partial}{\partial x} (3 x^{2} y^{4} + 1)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-ay^'=0 ?
The solution for y^{''}-ay^'=0 is y=c_{1}e^{at}+c_{2}