What is the solution for y^'-y=9te^2t,y(0)=1 ?

The solution for y^'-y=9te^2t,y(0)=1 is y=-9e^2(t^2+2(t+1))+(1+18e^2)e^t

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y^'-y=9te^2t,y(0)=1

y^{^{'}} - y = 9 t e^{2} t, y (0) = 1

y = - 9 e^{2} (t^{2} + 2 (t + 1)) + (1 + 18 e^{2}) e^{t}

Solution steps

y^{'} (t) - y = 9 t e^{2} t

Solve linear ODE:

y = - 9 e^{2} (t^{2} + 2 (t + 1)) + (1 + 18 e^{2}) e^{t}

y = - 9 e^{2} (t^{2} + 2 (t + 1)) + (1 + 18 e^{2}) e^{t}

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

\int_{0}^{\frac{π}{4}} cos^{4} (x) d x

d er i v a t i v e 25 x^{4} + 30 x^{3} - 11 x^{2} - 12 x + 4

(5 t + 3 y + 6) d t - d y = 0

y^{^{'}} - 3 y = 9 x

What is the solution for y^'-y=9te^2t,y(0)=1 ?
The solution for y^'-y=9te^2t,y(0)=1 is y=-9e^2(t^2+2(t+1))+(1+18e^2)e^t