IVP prob's: ydy/dx = x/sqrt1 + y2, y(0) = 1; etc.

G

Guest

Guest
I really need help with these two problems...

1. y dy/dx= x / sqrt1 + y^2, y(0) = 1

2. y'(x) = e^x + y, y(0) = 1
 
1)
let us get all y terms on one side of the = sign and x on the other
y dy/dx = x/[y^2-1]^1/2 multiply both sides by [y^2-1]^1/2
y[y^2-1]^1/2 dy/dx= x multiply both sides by dx
[y^2-1]^1/2 [ydy]= xdx the derivative of y^2 is 2y dy
2[y^2-1]^1/2 [2ydy]=x dx integrate
2[y^2-1]^3/2 [2/3] =x^2/2 + C
but x=0 y=1
4/3 [1-1]^3/2 = 0+C
C=0
multiply both sides by 6 [ I hate fractions]
8[y^2-1]^3/2 =3x^2 answer

please check for errors

I must think about 2
Arthur
 
2)
same approach as 1
dy/dx = e^(x+y)
dy/dx=e^xe^y
e^-y dy/dx=e^x
e^-y dy = e^x dx
- [e^-y][-dy]=e^x dx integrate
-e^-y=e^x +C
at x=0 y=1
-e^-1 =1+C
C=-[1+1/e]
C=-[e+1]/e substitute

-e^-y =e^x-[e+1]/e
1/e^y=[e+1]/e -e^x
1=e^(y-1)[e+1] -e^(x+y)

please check for errors
answer looks weird, but approach is right
Arthur
 
Brit-
Your instructor has probably been speaking of , "separation of variables" That is what we did for both problems
Arthur
 
Hello, Brit412!

If there are no parentheses in #2, more work is required.


\(\displaystyle \L2)\;\;\frac{dy}{dx} \:=\:e^x\,+\, y,\;\;y(0)\,= \,1\)

We have: \(\displaystyle \L\:\frac{dy}{dx}\,-\,y\:=\:e^x\)

Integrating factor: \(\displaystyle \:I \:=\:e^{\int(-1)dx} \:=\:e^{-x}\)

We have: \(\displaystyle \L\:e^{-x}\cdot\frac{dy}{dx}\,-\,e^{-x}\cdot y\;=\;e^{-x}\cdot e^x\)

Then: \(\displaystyle \L\:\frac{d}{dx}\left(e^{-x}\cdot y\right) \:=\:1\)

Integrate: \(\displaystyle \L\:e^{-x}\cdot y \;=\;x\,+\,C\)

Hence: \(\displaystyle \L\:y \;=\;e^x(x\,+\,C)\)


Since \(\displaystyle y(0)\,=\,1\), we have: \(\displaystyle \:1\:=\:e^0(0\,+\,C)\;\;\Rightarrow\;\;C\,=\,1\)

Therefore: \(\displaystyle \L\:y \;=\;e^x(x\,+\,1)\)

 
1. y dy/dx= x / sqrt1 + y^2, y(0) = 1

y dy/dx = x/[y^2-1]^1/2 multiply both sides by [y^2-1]^1/2

Possible typos:
Question 1: Should that "-1" be "+1"?

[y^2-1]^1/2 [ydy]= xdx the derivative of y^2 is 2y dy
2[y^2-1]^1/2 [2ydy]=x dx integrate

2nd question: Should that first "2" be a "1/2"?
 
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