Expert answers to "Solving ODE and finding maximaConsider the initial value..."

osteoblogda

Answered question

2022-01-20

Solving ODE and finding maxima
Consider the initial value problem

y^{'} + 5 y = 5 t, y (0) = y_{0}

. Find the value of

y_{0}

for which the solution touches, but does not cross, the t-axis.

Beverly Smith

Beginner2022-01-20Added 42 answers

First, your solution is incorrect.

y^{'} (t) + 5 y (t) = 5 t \Rightarrow \exp (5 t) y^{'} (t) + 5 y (t) \exp (5 t) = 5 t \exp (5 t)

\Rightarrow \frac{d (y (t) \exp (5 t))}{d t} = 5 t \exp (5 t)

Hence,

y (t) \exp (5 t) = (t - \frac{1}{5}) \exp (5 t) + c \Rightarrow y (t) = t - \frac{1}{5} + c \exp (- 5 t)

The slope is given by

y^{'} (t) = 5 (t - y (t)) = 5 (\frac{1}{5}, - c \exp (- 5 t)) = 1 - 5 c \exp (- 5 t)

For this to touch but not cross the t axis, we need

y (t_{0}) = 0 and y^{'} (t_{0}) = 0

, These give us

t_{0} - \frac{1}{5} + c \exp (- 5 t_{0}) = 0

and

1 - 5 c \exp (- 5 t_{0}) = 0

Now obtain

t_{0}

and c by solving these two equations. This gives us

t_{0} = 0 and c = \frac{1}{5}

. Hence, the curve is given by

y (t) = t - \frac{1}{5} + \frac{\exp (- 5 t)}{5}

.
Now

y_{0} = y (0) = 0

Tiefdruckot

Beginner2022-01-21Added 46 answers

y^{'} + 5 y = 5 t \to μ (t) = \exp (\int 5 d t) = e^{5 t}

\to d (e^{5 t} y) = 5 t e^{5 t} \to y = t - \frac{1}{5} + C e^{- 5 t}

y (0) = y_{0}

y = 25 t - 25 + (y_{0} + \frac{1}{5}) e^{- 5 t}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?