Problem with a differential equation

[Sga001]

Hello, i just did one problem, 2 different ways and got 2 different answers, that are not equal. I would appreciate if someone could point at any mistake i made, or explain the situation.

the problem says:

Find C given the initial condition.

y(x+1)+y' = 0 Initial Condition : y(-2) = 1

The book has the correct answer as the one on the right side.

Thank You for your help.

 

[galactus]

Your first one is appears to be correct.

\(\displaystyle \L\\y(x+1)+y'=0, \;\ y(-2)=1\)

Separate variables:

\(\displaystyle \L\\-x-1=\frac{dy}{y}\)

Integrate:

\(\displaystyle \L\\\frac{-x^{2}}{2}-x+C=ln|y|\)

\(\displaystyle \L\\e^{\frac{-x^{2}}{2}-x}e^{C}=y\)

$\displaystyle e^{C}=C_{1}$

\(\displaystyle \L\\C_{1}e^{\frac{-x^{2}}{2}-x}=y\)

Using the initial condition, we find that C_1 =1

\(\displaystyle \H\\y=e^{\frac{-x^{2}}{2}-x}\)

 

[Sga001]

The way i see it, is that this problem has 2 different answers, and the book only has one of them

 

[galactus]

No, your second approach has a mistake in the integration.

\(\displaystyle \L\\\int{(x+1)}dx\neq\frac{(x+1)^{2}}{2}\)

It's the same as you got in the other method.

 

[Sga001]

No, your second approach has a mistake in the integration.

\(\displaystyle \L\\\int{(x+1)}dx\neq\frac{(x+1)^{2}}{2}\)

It's the same as you got in the other method.

Hello, that is not a mistake, using the chain rule:

u = x+1
u' = 1

[(u^2)/2] * u'
[(x+1)^2)/2]*1 = \(\displaystyle \L\\\frac{(x+1)^{2}}{2}\) + C

 

[galactus]

The chain rule?. Are you differentiating or integrating?.

\(\displaystyle \L\\\int(x+1)dx=\frac{x^{2}}{2}+x+C\)

You could let C=1/2, I suppose.

 

[Sga001]

Hello, i apologize for what I said, im actually using the Antidifferentiation of a Composite Function


\(\displaystyle \int {f(u)du = F(u) + C}
\\)

In my case:


\(\displaystyle \[
\begin{array}{l}
f(u) = x + 1 \\
du = 1 \\
\end{array}
\]\)

Where F(u) is the antiderivative of f, in this case

\(\displaystyle \[
F(u) = \frac{{u^2 }}{2} + C
\]\)

 

[soroban]

Hello, Sga001!

This apparent contradiction occurs frequently in Calculus.
. . We must be constantly alert for it.


Example: \(\displaystyle \L\int(x + 1)\,dx\)


Method #1: \(\displaystyle \L\:\int x\,dx\,+\,\int 1\,dx \;=\;\frac{1}{2}x^2\,+\,x\,+\,C_1\)


Method #2: $\displaystyle \,\text{Let }u\:=\:x\,+\,1\;\;\Rightarrow\;\;du\,=\,dx$

Substitute: \(\displaystyle \L\:\int u\,du \;=\;\frac{1}{2}u^2\,+\,k\;=\;\frac{1}{2}(x\,+\,1)^2\,+\,k\)

. . . . . . . . \(\displaystyle \L= \:\frac{1}{2}x^2\,+\,x\,+\,\underbrace{\frac{1}{4}\,+\,k}\)


Since $\displaystyle \frac{1}{4}\,+\,k$ (one-quarter plus an arbitrary constant)
. . is another arbitrary constant $\displaystyle (C_2)$ , we have: \(\displaystyle \L\,\frac{1}{2}x^2\,+\,x\,+\,C_2\)
. . and, of course, $\displaystyle C_1\,=\,C_2$

But $\displaystyle C_1$ and $\displaystyle k$ are not independent.
. . They are related by: $\displaystyle C_1 \:=\:\frac{1}{4}\,+\,k$
This is a very subtle and evasive concept.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

In your problem, you integrated two ways and got two answers.

\(\displaystyle \text{Method #1: }\:\ln y \;=\;-\frac{1}{2}x^2\,-\,x\,+\,C_1\)

\(\displaystyle \text{Method #2: }\:\ln y \:=\:-\frac{1}{2}(x\,+\,1)^2\,+\,k \:=\:-\frac{1}{2}x^2\,-\,x\,\underbrace{-\,\frac{1}{4}\,+\,k}_{C_2}\)

Note: The original constants of integration, $\displaystyle C_1$ and $\displaystyle k$, are not independent.
. . . . .They are related: $\displaystyle C_1\:=\:-\frac{1}{4}\,+\,k\;\;\Rightarrow\;\;k\:=\:C_1\,+\,\frac{1}{4}$

So the second line should have looked like this:
. . \(\displaystyle \text{Method #2: }\;\ln y \:=\;-\frac{1}{2}(x\,+\,1)^2\,+\,C_1\,+\,\frac{1}{4}\)

We simply weren't aware of it.