Question

Find the solution of the differential equation y'=(5t+1)^2 satisfying the initial condition y(0)=4

Answer 1

Start by taking the integral of y':
\[\int\limits (5t+1)^2dt\]
Doing a u-sub,
u = 5t + 1
du = 5 dt
\[ \int\limits_{}{} u^2 (5)du\\ 5 \int\limits_{}{} u^2du \\ \frac{5}{3}u^3+C\]
So, if y= (5/3)(5t+1)+C, and y(0) = 4, what must C be?

Answer 2

Ah, beat me to it.

Answer 3

hold on

Answer 4

idk honestly how would you get that?

Answer 5

@bloopman can you help me?

Answer 6

\[\frac{ 5 }{ 3 }(5(0)+1)+C=4\\ \frac{5}{3}+C=4 \\ C = \frac{7}{3} \\. \\. \\. \\. \\ y = \frac{5}{3}(5x+1)+\frac{7}{3}\]

Answer 7

is that it or their is more beyond this?

Answer 8

I don't know what notation your answer needs to be in, but I would recommend simplifying that final equation.

Answer 9

the final answer has to be in "y=" form

Answer 10

in the instructions y' means find the first derivative

Answer 11

If the answer needs to be in y= form, just simplify the answer we got after solving for C.

Answer 12

25/3x+2/3?

Answer 13

is that the simplified version?

Answer 14

Woops Camp you missed a little something in your U-sub.
\[\Large du=5dt \qquad\to\qquad \frac{1}{5}du=dt\]

Answer 15

@DDCamp

Answer 16

@zepdrix where do we go from there?

Answer 17

So our substitution should actually give us:\[\Large \color{red}{\cancel{\color{black}{5 \int\limits u^2du}}}, \qquad\qquad\qquad \frac{1}{5}\int\limits u^2\;du\]Everything else looks ok though :o

Answer 18

Confused about any of the steps?

Answer 19

is it cool if you can go do the next steps ? :)

Answer 20

Integrating, applying the power rule for integration:\[\Large \int\limits y'\quad=\quad \frac{1}{5}\int\limits u^2\;du\]gives us:\[\Large y\quad=\quad \frac{1}{5}\cdot \frac{1}{3}u^3+C\]

Answer 21

\[\Large y(t)\quad=\quad \frac{1}{15}u^3+C\]Undoing our sub,\[\Large y(t)\quad=\quad \frac{1}{15}(5t+1)^3+C\]

Answer 22

\[\Large y(\color{#F35633}{t})\quad=\quad \frac{1}{15}(5\color{#F35633}{t}+1)^3+C\]

Answer 23

Using our initial data:\[\Large y(\color{#F35633}{0})\quad=\quad \frac{1}{15}(5\cdot\color{#F35633}{0}+1)^3+C\quad=\quad 4\]We can solve for C

Answer 24

Mmm so what do we get for C?

Answer 25

60/15

Answer 26

When we plug 0 in for t, we should still get a 1 being cubed inside of those brackets.

Answer 27

i meant 59/15

Answer 28

Oh ok :)

Answer 29

So we plug that in for our newly found C value,
\[\Large y(t)\quad=\quad \frac{1}{15}(5t+1)^3+\color{orangered}{\frac{59}{15}}\]

Answer 30

And we're done! yay

Answer 31

thanks you soo much you came to the rescue haha

Answer 32

:3