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 the general inseparable equation

[math]\displaystyle{ \frac{dy}{dx}+p(x)y=q(x) }[/math]

Now an integrating factor μ is defined as

[math]\displaystyle{ \mu = e^{\int p(x)\,dx} }[/math]

Thus:

[math]\displaystyle{ \frac{d\mu}{dx}=\left(e^{\int p(x)\,dx}\right)\frac{d}{dx} \left(\int p(x)\,dx\right) }[/math]

[math]\displaystyle{ \frac{d\mu}{dx}=\mu p(x) }[/math]

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

[math]\displaystyle{ \mu\frac{dy}{dx}+\mu p(x)y = \mu q(x) }[/math]

[math]\displaystyle{ \mu\frac{dy}{dx}+y\frac{d\mu}{dx}=\mu q(x) }[/math]

(using the product rule in reverse)

[math]\displaystyle{ \frac{d}{dx}(\mu y) = \mu \, q(x) }[/math]

[math]\displaystyle{ \mu y = \int \mu \, q(x)\,dx }[/math]

Finally, we obtain:

[math]\displaystyle{ 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, i.e., linear differential equations. For example, solving the inseparable equation:

[math]\displaystyle{ \frac{dy}{dx}=x+y }[/math]

[math]\displaystyle{ \frac{dy}{dx}-y=x }[/math]

By arranging in the form required, we obtain:

[math]\displaystyle{ p(x)=-1 }[/math]

[math]\displaystyle{ q(x)=x }[/math]

[math]\displaystyle{ \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]\displaystyle{ y = \frac{\int \mu \, q(x) \, dx}{\mu} }[/math].

[math]\displaystyle{ \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]\displaystyle{ y=\frac{\int xe^{-x}\,dx}{e^{-x}} }[/math]

[math]\displaystyle{ y = e^x \left(-xe^{-x} - e^{-x} + C\right) }[/math]

[math]\displaystyle{ y=Ce^{x}-x-1 }[/math]

Consider for example the inseparable equation

[math]\displaystyle{ 2y''+3y'+y=5. }[/math]

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

[math]\displaystyle{ \mathcal{L}\{f'\} = s \mathcal{L}\{f\} - f(0) }[/math]

[math]\displaystyle{ \mathcal{L}\{f''\} = s^2 \mathcal{L}\{f\} - s f(0) - f'(0) }[/math]

[math]\displaystyle{ \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]\displaystyle{ y(0)=0 }[/math] and [math]\displaystyle{ y'(0)=0. }[/math] Then,

[math]\displaystyle{ 2 \left(s^2 Y(s) - s\cdot 0 - 0 \right) + 3 \left(s Y(s) - 0\right) + Y(s) = \frac{5}{s}. }[/math]

It follows that

[math]\displaystyle{ \left(2s+1\right)\left(s+1\right) Y(s) = \frac{5}{s} }[/math]

or

[math]\displaystyle{ Y(s) = \frac{5}{s \left(2s+1\right) \left(s+1\right)}. }[/math]

Now one can just take the inverse Laplace transform of Y(s) to get the solution y to the original equation.

See also

• Differential equation
• Examples of differential equations
• Numerical methods for ordinary differential equations
• Method of variation of parameters