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

karush

Well-known member

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
$$(1+y^2)\, dy=(2+2x)\, dx$$

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MarkFL

Staff member
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

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

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MarkFL

Staff member
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

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

there is no book answer to this

ok sorry im kinda lost

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skeeter

Well-known member
MHB Math Helper
$\arctan(y) = (1+x)^2 + C$

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

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

karush

Well-known member
how did you get..
 $(1+x)^2$

topsquark

Well-known member
MHB Math Helper
so....then,,,,
$$\arctan \left(y\right)=x^2+2x$$
then
$$y=\tan(x^2+2x)$$

there is no book answer to this

ok sorry im kinda lost
You aren't lost. You got the correct answer!

-Dan

skeeter

Well-known member
MHB Math Helper
$\displaystyle \int 2(1+x) \, dx = (1+x)^2 +C$

MarkFL

$$\displaystyle \int_0^y \frac{1}{u^2+1}\,du=2\int_0^x v+1\,dv$$
$$\displaystyle \int_0^y \frac{1}{u^2+1}\,du=2\int_1^{x+1} w\,dw$$
$$\displaystyle \arctan(y)=(x+1)^2-1=x(x+2)$$
$$\displaystyle y=\tan(x(x+2))$$