What is the solution for 1/(2*5)y+y^'= t/2 ,y(30)=10 ?

The solution for 1/(2*5)y+y^'= t/2 ,y(30)=10 is y=5(t-10)-90e^{(30-t)/(10)}

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1/(2*5)y+y^'= t/2 ,y(30)=10

\frac{1}{2 \cdot 5} y + y^{^{'}} = \frac{t}{2}, y (30) = 10

y = 5 (t - 10) - 90 e^{\frac{30 - t}{10}}

Solution steps

\frac{1}{2 \cdot 5} y + y^{'} (t) = \frac{t}{2}

Solve linear ODE:

y = 5 (t - 10) - 90 e^{\frac{30 - t}{10}}

y = 5 (t - 10) - 90 e^{\frac{30 - t}{10}}

(- 10, 1), (0, - 4)

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x \to π lim (sin (\frac{7}{6} x))

\int \frac{1}{e ^{x + 1}} d x

What is the solution for 1/(2*5)y+y^'= t/2 ,y(30)=10 ?
The solution for 1/(2*5)y+y^'= t/2 ,y(30)=10 is y=5(t-10)-90e^{(30-t)/(10)}