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Popular >

t^2y^{''}-ty^'+y=0

Solution

t^{2} y^{^{''}} - t y^{^{'}} + y = 0

Solution

y = c_{1} t + c_{2} ln (t) t

Solution steps

t^{2} y^{''} (t) - t y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} t + c_{2} ln (t) t

y = c_{1} t + c_{2} ln (t) t

Graph

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Frequently Asked Questions (FAQ)

What is the solution for t^2y^{''}-ty^'+y=0 ?
The solution for t^2y^{''}-ty^'+y=0 is y=c_{1}t+c_{2}ln(t)t