-b.2.2.26 Solve first order IVP and determine where minimum of solution occurs

[karush]

OK going to comtinue with these till I have more confidence with it
$$\dfrac{dy}{dx}=2 (1+x) (1+y^2), \qquad y(0)=0$$
separate
$$(1+y^2)\, dy=(2+2x)\, dx$$

 

[MarkFL]

The ODE associated with this IVP is separable. I would next write:

\(\displaystyle \int_0^y \frac{1}{u^2+1}\,du=2\int_0^x v+1\,dv\)

And...GO!

 

[karush]

why would this need to be a u=v substitution?
do you just plug in y=0, x=0

 

[MarkFL]

I changed the dummy variables of integration because I used the boundaries as the limits of the definite integrals. It's considered bad form to have the variable of integration in the limits. Using definite integrals removes the need for finding the constant of integration.

Suppose you have the initial value problem (IVP):

\(\displaystyle \frac{dy}{dx}=f(x)\) where \(\displaystyle y\left(x_0\right)=y_0\)

Now, separating variables and using indefinite integrals, we may write:

\(\displaystyle \int\,dy=\int f(x)\,dx\)

And upon integrating, we find

\(\displaystyle y(x)=F(x)+C\) where \(\displaystyle \frac{d}{dx}\left(F(x) \right)=f(x)\)

Using the initial condition, we get

\(\displaystyle y\left(x_0 \right)=F\left(x_0 \right)+C\)

Solving for \(C\) and using \(\displaystyle y\left(x_0\right)=y_0\), we obtain:

\(\displaystyle C=y_0-F\left(x_0 \right)\) thus:

\(\displaystyle y(x)=F(x)+y_0-F\left(x_0 \right)\)

which we may rewrite as:

\(\displaystyle y(x)-y_0=F(x)-F\left(x_0 \right)\)

Now, we may rewrite this, using the anti-derivative form of the fundamental theorem of calculus, as:

\(\displaystyle \int_{y_0}^{y(x)}\,dy=\int_{x_0}^{x}f(x)\,dx\)

Now, since the variable of integration gets integrated out, it is therefore considered a "dummy variable" and since it is considered good form not to use the same variable in the boundaries as we use for integration, we may switch these dummy variables and write:

\(\displaystyle \int_{y_0}^{y(x)}\,du=\int_{x_0}^{x}f(v)\,dv\)

This demonstrates that the two methods are equivalent.

Using the boundaries (the initial and final values) in the limits of integration eliminates the need to solve for the constant of integration, and I find it a more intuitive and cleaner approach to separable initial value problems.

 

[karush]

so...then,,,,
$$\arctan \left(y\right)=x^2+2x$$
then
$$y=\tan(x^2+2x)$$

there is no book answer to this

ok sorry I am kinda lost

 

[skeeter]

$\arctan(y) = (1+x)^2 + C$

$y(0) = 0 \implies C = -1$

$y = \tan(x^2+2x)$

 

[karush]

how did you get..

$(1+x)^2 $

 

[topsquark]

so...then,,,,
$$\arctan \left(y\right)=x^2+2x$$
then
$$y=\tan(x^2+2x)$$

there is no book answer to this

ok sorry I am kinda lost

You aren't lost. You got the correct answer!

-Dan

 

[skeeter]

$\displaystyle \int 2(1+x) \, dx = (1+x)^2 +C$

 

[MarkFL]

The ODE associated with this IVP is separable. I would next write:

\(\displaystyle \int_0^y \frac{1}{u^2+1}\,du=2\int_0^x v+1\,dv\)

And...GO!

Continuing, we have:

\(\displaystyle \int_0^y \frac{1}{u^2+1}\,du=2\int_1^{x+1} w\,dw\)

\(\displaystyle \arctan(y)=(x+1)^2-1=x(x+2)\)

\(\displaystyle y=\tan(x(x+2))\)