What is the solution for 33.62x^'+x=100 ?

The solution for 33.62x^'+x=100 is x=-e^{(-50t-c_{1})/(1681)}+100

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33.62x^'+x=100

33.62 x^{^{'}} + x = 100

x = - e^{\frac{- 50 t - c _{1}}{1681}} + 100

Solution steps

33.62 x^{'} (t) + x = 100

Solve separable ODE:

x = - e^{\frac{- 50 t - c _{1}}{1681}} + 100

x = - e^{\frac{- 50 t - c _{1}}{1681}} + 100

7 e^{- x^{2}}

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + e^{- x}

y = \frac{x ^{2}}{( x + 2 )}

ln (x^{8})

\int \frac{sin ^{2} ( θ )}{cos ^{4} ( θ )} d θ

What is the solution for 33.62x^'+x=100 ?
The solution for 33.62x^'+x=100 is x=-e^{(-50t-c_{1})/(1681)}+100