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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

- Thread starter karush
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- Jan 31, 2012

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- #3

- Jan 31, 2012

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why would this need to be a u=v substitution?

do you just plug in y=0, x=0

do you just plug in y=0, x=0

Last edited:

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- #4

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.

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- #5

- Jan 31, 2012

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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

$$\arctan \left(y\right)=x^2+2x$$

then

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

there is no book answer to this

ok sorry im kinda lost

Last edited:

- Mar 1, 2012

- 854

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

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

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

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

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

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- #7

- Jan 31, 2012

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how did you get..

$(1+x)^2 $ |

- Aug 30, 2012

- 1,180

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

-Dan

- Mar 1, 2012

- 854

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

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- #10

Continuing, we have: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!!

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

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

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