Inseparable differential equation

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, etc.

Examples

Consider for example the inseparable equation

:$2y"+3y\text{'}+y=5.$

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

: $mathcal\{L\}\{f\text{'}\}\; =\; s\; mathcal\{L\}\{f\}\; -\; f(0)$: $mathcal\{L\}\{f"\}\; =\; s^2\; mathcal\{L\}\{f\}\; -\; s\; f(0)\; -\; f\text{'}(0)$: $mathcal\{L\}left\{\; f^\{(n)\}\; ight\}\; =\; s^n\; mathcal\{L\}\{f\}\; -\; s^\{n\; -\; 1\}\; f(0)\; -\; cdots\; -\; f^\{(n\; -\; 1)\}(0).$

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 $y(0)=0$ and $y\text{'}(0)=0.$ Then,

: $2(s^2Y-scdot\; 0-0)+3(s\; Y-0)+Y=frac\{5\}\{s\}.$It follows that:$(2s+1)(s+1)Y=frac\{5\}\{s\}$ or:$Y=frac\{5\}\{s(2s+1)(s+1)\}.$Now one can just take the inverse Laplace transform of "Y" to get the solution "y" to the original equation.

ee also

* examples of differential equations
* method of variation of parameters