Question

If dy/dx = x * square root of (2x^2+1) , and y contains the point (2,4), then the y-intercept of the particular solution is A)0 B) -2/3 C)-1/3 D)-4/3 or E)-1?

Answer 1

\[d/x=x \sqrt{2x ^{2}+1}\]

Answer 2

There is the equation in an easier format!

Answer 3

do you know how to integrate?

Answer 4

Yeah I got the integral of the equation, I'm just not sure how to find the y-intercept!

Answer 5

can i see what you after you integrated

Answer 6

we can find the constant of integration by using the point (2,4)
then we can find the y-intercept by putting x=0 and solving for y

Answer 7

\[1/6(2x^2+1)^{2/3}+c\]

Answer 8

I think you mean the power to be 3/2 right?
anyways solve this for C first
\[y=\frac{1}{6}(2x^2+1)^\frac{3}{2}+C \\ 4=\frac{1}{6}(2(2)^2+1)^\frac{3}{2}+C\]

Answer 9

Oops yeah I did!

Answer 10

I got c=1/2

Answer 11

oops c=-1/2

Answer 12

\[4=\frac{1}{6}(2(4)+1)^\frac{3}{2}+C \\ 4=\frac{1}{6}(8+1)^\frac{3}{2}+C \\ 4=\frac{1}{6}(9)^\frac{3}{2}+C \\ 4=\frac{1}{6}(3)^3+C \\ 4=\frac{1}{6}(27)+C\]
yeah should be -1/2 :)

Answer 13

\[y=\frac{1}{6}(2x^2+1)^\frac{3}{2}-\frac{1}{2}\]

Answer 14

to find the y-intercept set x=0 and solve for y

Answer 15

I got -1/3! So the answer wold be C

Answer 16

Can I ask you one more question possibly?

Answer 17

sounds great

Answer 18

sure

Answer 19

Okay so the question is "Solve each initial value problem."
The problem is: \[dy/dx=e ^{x-y}\] and f(0)=2

Answer 20

first you know law of exponents
so you know x^(A+B)
can be written as x^A times x^B

Answer 21

\[\frac{dy}{dx}=e^{x}e^{-y}\]

Answer 22

Yep!

Answer 23

no separate your variables

Answer 24

\[e^{y} dy=e^{x} dx\]
integrate both sides

Answer 25

I'm not sure how you integrate with these

Answer 26

\[\int\limits_{}^{}e^u du=e^u+C\]

Answer 27

since d(e^u)/du=e^u

Answer 28

\[\int\limits_{}^{}e^y dy=\int\limits e^x dx \\ e^y+C_1=e^x+C_2 \\ e^y=e^x+C_2-C_1 \\ e^y=e^x+C\]

Answer 29

you can solve for y if you want to

Answer 30

you are given y(0)=2
this means replace x with 0 and y with 2
so you can solve for C

Answer 31

when i substituted for y and x I got c to be \[e ^{2}-1\]

Answer 32

sounds awesome so pluggin that back into our solution we have
\[e^y=e^x+e^2-1\]

Answer 33

you can leave like this
but sometimes it is preferred to solve for y

Answer 34

how would I solve for y?

Answer 35

do you know the inverse function of f(x)=e^x is g(x)=ln(x)

Answer 36

yep!

Answer 37

take ln( ) of both sides

Answer 38

\[e^y=e^x+e^2-1 \\ \ln(e^y)=\ln(e^x+e^2-1) \\ y \ln(e)=\ln(e^x+e^2-1) \\ y(1)=\ln(e^x+e^2-1) \\ y=\ln(e^x+e^2-1)\]

Answer 39

Great! Thank you so much!
I also had a problem that was finding the initial value and the equation was \[dy/dx=xe ^{y}\] and f(0)=0 and I ended up with \[(x ^{2}*e ^{y})/2 - 1/2\] is that right?

Answer 40

why no equal sign anywhere?

Answer 41

the second equation is equal to y, just forgot to add it sorry

Answer 42

ok well this is another one you can do by seperation of variables
\[\frac{dy}{dx}=xe^y \\ dy=x e^y dx \\ e^{-y} dy=x dx\]
and then you integrate both sides

Answer 43

\[\int\limits_{}^{}e^{-y} dy=\int\limits x dx \\ -e^{-y}=\frac{x^2}{2}+C\]

Answer 44

use y(0)=0 to find your constant of integration

Answer 45

alright i did that so c=1

Answer 46

-1?

Answer 47

yes!

Answer 48

yep sounds good plug into our solution

Answer 49

\[-e^{-y}=\frac{x^2}{2}-1\]
you can leave it as this or solve for y

Answer 50

you can solve for y in a very similar way we did before

Answer 51

your solutions will not always be easy or could be impossible to solve for y

Answer 52

is this one of those problems?

Answer 53

no it isn't
this is is easy :p

Answer 54

you multiply -1 on both sides
then take ln( ) of both sides

Answer 55

and one more step remains after

Answer 56

would it be \[y=\ln ((x ^{2}/2)+1)\]

Answer 57

close but still off

Answer 58

alright what else do I have to do?

Answer 59

\[-e^{-y}=\frac{x^2}{2}-1 \\ \text{ multiply -1 on both sides } \\ e^{-y}=1-\frac{x^2}{2}\]
you cool with that step?

Answer 60

yep!

Answer 61

take ln( ) of both sides

Answer 62

\[\ln(e^{-y})=\ln(1-\frac{x^2}{2})\]

Answer 63

we can use power rule for logs on the left hand side

Answer 64

that is you can do this:
\[-y \ln(e)=\ln(1-\frac{x^2}{2})\]

Answer 65

ln(e)=1 though

Answer 66

\[-y=\ln(1-\frac{x^2}{2})\]
the last step that remains is multiplying both sides by -1

Answer 67

\[y=- \ln(1-\frac{x^2}{2})\]

Answer 68

there is something else you can do but we don't have to go too crazy

Answer 69

unless you were trying to match an answer to a choice you had or something

Answer 70

No I'm not matching it to a choice, what else can we do?

Answer 71

you can use power rule again

Answer 72

\[y=- \ln(\frac{2-x^2}{2}) \\ y=\ln((\frac{2-x^2}{2})^{-1}) \\ y=\ln(\frac{2}{2-x^2})\]

Answer 73

I think that looks pretty but you could also use the quotient rule and write it like this
\[y=\ln(2)-\ln(2-x^2)\]

Answer 74

That does look better!

Answer 75

Great! Thank you so much for all of your help!!

Answer 76

np :)