en>Tomruen at 03:32, 7 November 2013

2013-11-07T03:32:51Z

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	~~The name~~ of the ~~writer is Numbers~~. ~~Years ago~~ we ~~moved~~ to ~~North Dakota and I love each working day residing right here~~. ~~Doing ceramics~~ is ~~what my family members~~ and ~~I enjoy~~. ~~For years he~~'s ~~been working as~~ a ~~meter reader~~ and it'~~s some thing he really enjoy~~.<br><br>~~My web blog~~ - ~~[http~~://~~dateplate~~.~~com/dayzfe std testing at home~~]		{{Unreferenced\|date=December 2009}}
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			In [[mathematics]], an '''inseparable differential equation''' is an [[ordinary differential equation]] that cannot be solved by using [[separation of variables]]. To solve an inseparable differential equation one can employ a number of other methods, like the [[Laplace transform]], [[substitution rule\|substitution]], etc.

			==Examples==
			Consider the general inseparable equation

			:<math> \frac{dy}{dx}+p(x)y=q(x)</math>

			Now we will define a special factorial, ''μ'' as

			:<math>\mu = e^{\int p(x)dx}</math>

			Thus:

			:<math>\frac{d\mu}{dx}=(e^{\int p(x)dx})\frac{d}{dx} (\int p(x)dx)</math>
			:<math>\frac{d\mu}{dx}=\mu p(x)</math>

			From here we can solve the equation using the above definition:

			:<math>\mu\frac{dy}{dx}+\mu p(x)y = \mu q(x)</math>
			:<math>\mu\frac{dy}{dx}+y\frac{d\mu}{dx}=\mu q(x)</math>

			(using the product rule in reverse)

			:<math>\frac{d}{dx}(\mu y) = \mu q(x)</math>
			:<math>\mu y = \int \mu q(x)dx</math>

			Finally, we obtain:

			:<math>y=\frac{\int \mu q(x)dx}{\mu}</math>

			This can be used to solve most all inseparable equations containing no ''y'' to a degree other than one. For example, solving the inseparable equation:

			:<math>\frac{dy}{dx}=x+y</math>
			:<math>\frac{dy}{dx}-y=x</math>

			By arranging in the form required, we obtain:

			:<math> p(x)=-1\ </math>
			:<math> q(x)=x\ </math>
			:<math>\frac{dy}{dx}+p(x)y=q(x)</math>

			Now all that is necessary is to find the value of ''μ'' to plug into our original equation of <math>y=\frac{\int\mu q(x)dx}{\mu}.</math>

			:<math> \mu =e^{\int p(x)dx}=e^{\int -1 dx}=e^{-x}</math>

			Plugging this into the original equation and simplifying gives us our final answer:

			:<math> y=\frac{\int xe^{-x}}{e^{-x}}</math>
			:<math> y=e^x(-xe^{-x}-e^{-x}+C)\ </math>
			:<math> y=Ce^{x}-x-1\ </math>

			Consider for example the inseparable equation

			:<math>2y''+3y'+y=5.\ </math>

			Let us solve it using the Laplace transform. One has that

			: <math>\mathcal{L}\{f'\}
			= s \mathcal{L}\{f\} - f(0)</math>
			: <math>\mathcal{L}\{f''\}
			= s^2 \mathcal{L}\{f\} - s f(0) - f'(0)</math>
			: <math>\mathcal{L}\left\{ f^{(n)} \right\}
			= s^n \mathcal{L}\{f\} - s^{n - 1} f(0) - \cdots - f^{(n - 1)}(0).</math>

			Using the convenience that Laplace transforms follow the rules of linearity, one can solve the above example for ''y'' by performing a Laplace transform on both sides of the differential equation, substituting in the initial values, solving for the transformed function, and then performing an inverse transform.

			For the above example, assume initial values are <math>y(0)=0</math> and <math>y'(0)=0.</math> Then,

			: <math>2(s^2Y-s\cdot 0-0)+3(s Y-0)+Y=\frac{5}{s}. </math>
			It follows that
			:<math>(2s+1)(s+1)Y=\frac{5}{s}</math>
			or
			:<math>Y=\frac{5}{s(2s+1)(s+1)}.</math>
			Now one can just take the inverse Laplace transform of ''Y'' to get the solution ''y'' to the original equation.

			==See also==
			* [[examples of differential equations]]
			* [[method of variation of parameters]]

			{{DEFAULTSORT:Inseparable Differential Equation}}

			[[Category:Ordinary differential equations]]

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2012-07-30T20:34:08Z

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The name of the writer is Numbers. Years ago we moved to North Dakota and I love each working day residing right here. Doing ceramics is what my family members and I enjoy. For years he's been working as a meter reader and it's some thing he really enjoy.<br><br>My web blog - [http://dateplate.com/dayzfe std testing at home]

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en>Tomruen at 03:32, 7 November 2013

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