What is the solution for y^{''}-5y^'=25t,y(0)=5,y^'(0)=0 ?

The solution for y^{''}-5y^'=25t,y(0)=5,y^'(0)=0 is y=(5-24/5)e^{5t}+24/5-(5t^2)/2-t

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y^{''}-5y^'=25t,y(0)=5,y^'(0)=0

y^{^{''}} - 5 y^{^{'}} = 25 t, y (0) = 5, y^{^{'}} (0) = 0

y = (5 - \frac{24}{5}) e^{5 t} + \frac{24}{5} - \frac{5 t ^{2}}{2} - t

Solution steps

y^{''} (t) - 5 y^{'} (t) = 25 t

Solve linear ODE:

y = (5 - \frac{24}{5}) e^{5 t} + \frac{24}{5} - \frac{5 t ^{2}}{2} - t

y = (5 - \frac{24}{5}) e^{5 t} + \frac{24}{5} - \frac{5 t ^{2}}{2} - t

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What is the solution for y^{''}-5y^'=25t,y(0)=5,y^'(0)=0 ?
The solution for y^{''}-5y^'=25t,y(0)=5,y^'(0)=0 is y=(5-24/5)e^{5t}+24/5-(5t^2)/2-t