Examples of differential equations: Difference between revisions

Revision as of 13:59, 8 February 2021

Revision as of 14:15, 8 February 2021

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

The method for solving this equation relies on a special integrating factor, {{math|1=''&mu;μ'' = ''μ''(''x'')}}:<!--signify integrating factors should only be dependent on ''x''-->

:<math>\mu = e^{\int_{x_0}^x p(t)\, dt}</math>

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We choose this integrating factor because it has the special property that its derivative is itself times the function we are integrating, that is:

:<math>\frac{d{\mu}}{dx} = e^{\int_{x_0}^x p(t)\, dt} \cdot p(x) = \mu p(x)</math>

Multiply both sides of the original differential equation by ''&mu;''{{mvar|μ}} to get:

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

Because of the special ''&mu;''{{mvar|μ}} we picked, we may substitute {{math|{{sfrac|''d&mu;dμ''/|''dx''}}}} for {{math|''&mu;''&nbsp;''μ p''(''x'')}}, simplifying the equation to:

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

Using the [[product rule (calculus)|product rule]] in reverse, we get:

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

Integrating both sides with respect to {{mvar|x}}:

:<math>\mu{ y} = \left(C + \int\mu q(x)\, dx\right) + C</math>

Finally, to solve for ''{{mvar|y''}} we divide both sides by <math>\mu</math>{{mvar|μ}}:

:<math>y = \frac{\left(C + \int\mu q(x)\, dx\right) + C}{\mu}</math>

Since ''&mu;''{{mvar|μ}} is a function of ''{{mvar|x''}}, we cannot simplify any further directly.

==Second-order linear ordinary differential equations==