Question

Definite integral

Answer 1

\[\dfrac{dy}{dx} = e^x\]
y = 4 when x = 0

Answer 2

Just need a quick refresher on how to do these

Answer 3

@SolomonZelman

Answer 4

Do what?

Answer 5

Definite integrals... We did something slightly different in class with these

Answer 6

What is your integral?

Answer 7

is that your problem?

Answer 8

Yes

Answer 9

I know we have to figure out what C is with these, but forgot how to get the integral symbol in there

Answer 10

\int

Answer 11

Just by looking at it I know c is 3, but that's because e^0 is 1...

Answer 12

and with \displaystyle before \int you can type the integrals in a formatted (not squeezed) way even with `\( \)`

Answer 13

This is the whole thing with instructions

Answer 14

yes, I figured it is a DE problem:)

Answer 15

I will give you an example of a similar equation, so that you can obtain a procedure of it.

Answer 16

Okay, I also have \(\dfrac{dy}{dx} = \dfrac{x^2 + x + 1}{x}\) that I have to do if we want to just do that one

Answer 17

y = 0 when x = 1

Answer 18

\(\color{black}{ \displaystyle \frac{dy}{dx}=\frac{2x^2-3x+1}{x} }\)

you can integrate both sides with respect to x.

\(\color{black}{ \displaystyle \int \frac{dy}{dx}dx=\int \frac{2x^2-3x+1}{x}dx }\)

the differentials cancel on LHS

\(\color{black}{ \displaystyle \int dy=\int \frac{2x^2-3x+1}{x}dx }\)

you can integrate as follows
\(\color{black}{ \displaystyle \int \frac{2x^2-3x+1}{x}dx =\int \left(2x-3+\frac{1}{x} \right)dx=x^2-3x+\ln|x|+C_2 }\)
and the integral on LHS is obvious, \(\color{black}{ \displaystyle \int dy=y+C_1 }\)

thus you have
\(\color{black}{ \displaystyle y+C_1=x^2-3x+\ln|x|+C_2 }\)

arbitrary constants when subtracted or added still give arbitrary constant ..
\(\color{black}{ \displaystyle y=x^2-3x+\ln|x|+C }\) this is the general solution.

you are given that \(\color{black}{ \displaystyle y=2 }\) when \(\color{black}{ \displaystyle x=1 }\)

\(\color{black}{ \displaystyle 2=1^2-3(1)+\ln|(1)|+C }\)
\(\color{black}{ \displaystyle C=4 }\)

So,
\(\color{red}{ \displaystyle y=x^2-3x+\ln|x|+4 }\).

Answer 19

I changed the numbers a little, but basically everything is just the same.

Answer 20

As regards to your initial equation, I bet you can solve it in your head
(integrate both sides with respect to y, cancel the differentials, you know what this integration will give you, and use the point (1,4) to solve for C ....)

Answer 21

Anyway ... do you have any questions regarding the above example (or anything else in general) ? (Take your time if you need to.)

Answer 22

I think I got it... thanks so much for all your help tonight

Answer 23

No problem:)

Answer 24

I will do one more example (bonus) in a min or two.

Answer 25

\(\color{black}{ \displaystyle \frac{dy}{dx}=-8x^3e^y;~~~y(0)=\ln(8) }\)
----------------------

You already feel that in the future we are about to integrate both sides with respect to x, and then the LHS will be integrated with respect to y (since dx and dx will cancel), and RHS will be integrated with respect to x. So you would like to move anything with y to LHS.

\(\color{black}{ \displaystyle -e^{-y}\frac{dy}{dx}=8x^3}\)

then integrate with respect to x

\(\color{black}{ \displaystyle \int -e^{-y}\frac{dy}{dx}dx=\int 8x^3dx}\)

differentials cancel

\(\color{black}{ \displaystyle \int -e^{-y}dy=\int 8x^3dx}\)

integrate each side individually,

\(\color{black}{ \displaystyle e^{-y}+C_1=2x^4+C_2}\)

subtracting arbitrary constant ...

\(\color{black}{ \displaystyle e^{-y}=2x^4+C}\)

we need to solve for y,

\(\color{black}{ \displaystyle y=\ln|2x^4+C|}\) (this is the general solution)

given your initial condition, \(\color{black}{ \displaystyle y(0)=\ln(8)}\), we have:

\(\color{black}{ \displaystyle \ln(8)=\ln|2(0)^4+C|}\)
\(\color{black}{ \displaystyle \ln(8)=\ln|C|}\)
\(\color{black}{ \displaystyle C=8}\)

so your solution is:
\(\color{black}{ \displaystyle y=\ln|2x^4+8|}\)

Note that \(2x^4+11\) is always positive anyway, so you can rewrite without abs value,

\(\color{red}{ \displaystyle y=\ln(2x^4+8)}\).

Answer 26

Oh by the way, there is a mistake.

Answer 27

Where I have \(\color{black}{\displaystyle \color{black}{ \displaystyle e^{-y}=2x^4+C}}\) it is correct, and then ...
------------------------------------------------
Raise both sides to the power of \({-1}\).
\(\color{black}{\displaystyle \color{black}{ \displaystyle e^{-y}=2x^4+C}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note on the domain. You know that \(e^{-y}>0\) for all \(y\in \mathbb{R}\).
Thus you also know that \(\color{}{\displaystyle 2x^4+C>0}\).
\(\large (\)B/c, \(e^{-y}=2x^4+C\), and \(e^{-y}>0\).\(\large)\)
Thus, we may raise both sides to the power of \({-1}\).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Raising both sides to the power of \(-1\), gives:
\(\color{black}{\displaystyle \color{black}{ \displaystyle e^{y}=(2x^4+C)^{-1}}}\)

Then, since in general \(\color{black}{\displaystyle e^a=b\quad \Longleftrightarrow \quad a=\ln(b)}\)
(and this is achivied by taking the natural log of both sides),
therefore, in our case we have:

\(\color{black}{\displaystyle \displaystyle y=\ln[(2x^4+C)^{-1}]}\) \(\tiny \\[0.5em]\)
\(\color{black}{\displaystyle \displaystyle y=-\ln(2x^4+C)}\)

Then, \(y(0)=\ln(8)\), and thus:

\(\color{black}{\displaystyle \displaystyle \ln(8)=-\ln(2(0)^4+C)}\) \(\tiny \\[0.5em]\)
\(\color{black}{\displaystyle \displaystyle \ln(8)=-\ln(C)}\) \(\tiny \\[0.5em]\)
\(\color{black}{\displaystyle \displaystyle \ln(8)=\ln(1/C)}\) \(\tiny \\[0.5em]\)
\(\color{black}{\displaystyle \displaystyle C=1/8}\)

thus,
\(\color{black}{\displaystyle \displaystyle y=-\ln\left(2x^4+\frac{1}{8}\right)}\).

Answer 28

Or, you might as well, plug in the initial condition to solve for C, before when you had
\(\color{black}{\displaystyle \displaystyle e^{-y}=2x^4+C}\)
\(\color{black}{\displaystyle \displaystyle e^{-\ln(8)}=2(0)^4+C}\)
\(\color{black}{\displaystyle \displaystyle e^{\ln(1/8)}=C}\)
\(\color{black}{\displaystyle \displaystyle 1/8=C}\) (you get the same \(C\)).

the last step is due to the rule, \(\color{black}{\displaystyle e^{\ln f(x)}=f(x)}\).