Solution to Differential Equation/Examples/Arbitrary Order 2 ODE

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Examples of Solutions to Differential Equations

Consider the real function defined as:

$y = \map f x = \ln x + x$

defined on the domain $x \in \R_{>0}$.


Then $\map f x$ is a solution to the second order ODE:

$(1): \quad x^2 y'' + 2 x y' + y = \ln x + 3 x + 1$

defined on the domain $x \in \R_{>0}$.


Proof

It is noted that $\map f x$ is not defined in $\R$ when $x \le 0$ because that is outside the domain of the natural logarithm $\ln$.

The same constraint applies to $(1)$.


Having established that, we continue:

\(\ds y\) \(=\) \(\ds \ln x + x\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds \dfrac 1 x + 1\) Derivative of Natural Logarithm, Derivative of Identity Function
\(\ds \leadsto \ \ \) \(\ds y''\) \(=\) \(\ds -\dfrac 1 {x^2} + 0\) Power Rule for Derivatives, Derivative of Constant


Then:

\(\ds \) \(\) \(\ds x^2 \paren {-\dfrac 1 {x^2} } + 2 x \paren {\dfrac 1 x + 1} + \ln x + x\) substituting for $y$, $y'$ and $y''$ from above into the left hand side of $(1)$
\(\ds \) \(=\) \(\ds -1 + 2 + 2 x + \ln x + x\)
\(\ds \) \(=\) \(\ds \ln x + 3 x + 1\) which equals the right hand side of $(1)$

$\blacksquare$


Sources