Find y as a function of t, if dydt+(0.2)ty=4tandy(0)=9

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High School
Geometry
Trigonometry
Trigonometric Functions

Dottie Parra

Answered question

2021-09-14

Find y as a function of t, if

\frac{d y}{d t} + (0.2) t y = 4 t and y (0) = 9

Answer & Explanation

aprovard

Skilled2021-09-15Added 94 answers

Simplify $(\frac{d y}{d t}) + (0.2) t y = 4 t$ with $y (0) = 9$ :
$\frac{d y}{d t} + (0.2) t y = 4 t$
$\frac{d y}{d t} = 4 t - (0.2) t y$
$\frac{d y}{d t} = t (4 - 0.2 y)$
$\frac{d y}{4 - 0.2 y} = t d t$
$\ln t \frac{d y}{4 - 0.2 y} = \ln d t$
$\frac{\ln (4 - 0.2 y)}{- 0.2} = \frac{t^{2}}{2} + C$
$\ln (4 - 0.2 y) = - 0.1 t^{2} - 0.2 C$
$4 - 0.2 y = e^{- 0.1 t^{2} - 0.2 C}$
$0.2 y = 4 - e^{- 0.1 t^{2} - 0.2 C}$
$y = 20 - 5 e^{- 0.1 t^{2} - 0.2 C}$
Substitute $y (0) = 9$ :
$9 = 20 - 5 e^{- 0.1 t^{2} - 0.2 C}$
$e^{- 0.2 C} = \frac{11}{5}$
$- 0.2 C = \ln (\frac{11}{5})$
$y = 20 - 5 e^{- 0.1 t^{2} + \ln (\frac{11}{5})}$
$y = 20 - 5 \times \frac{11}{5} e^{- 0.1 t^{2}}$
$y = 20 - 11 e^{- 0.1 t^{2}}$
The answer:
$y = 20 - 11 e^{- 0.1 t^{2}}$

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