What is the solution for q^'+1/50 q=0.2,q(0)=5 ?

The solution for q^'+1/50 q=0.2,q(0)=5 is q(t)=-5e^{-t/(50)}+10

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q^'+1/50 q=0.2,q(0)=5

q^{^{'}} + \frac{1}{50} q = 0.2, q (0) = 5

q (t) = - 5 e^{- \frac{t}{50}} + 10

Solution steps

q^{'} (t) + \frac{1}{50} q (t) = 0.2

Solve separable ODE:

q (t) = - 5 e^{- \frac{t}{50}} + 10

q (t) = - 5 e^{- \frac{t}{50}} + 10

sin (x) sin (y) y^{^{'}} + cos (y) cos (x) = 0

3 x sin (x) d x + cos (y) d y = 0

y^{^{'}} = \frac{6 sin ( x )}{sin ( y )}, y (0) = \frac{π}{2}

y^{^{'}} = 2 x y^{2}, y (0) = \frac{1}{( 4 - x ^{2} )}

(x - 8)^{2} \frac{d y}{d x} = (7 - y)^{2}

What is the solution for q^'+1/50 q=0.2,q(0)=5 ?
The solution for q^'+1/50 q=0.2,q(0)=5 is q(t)=-5e^{-t/(50)}+10