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A linear differential equation of second or higher order have more than $1$ solutions of $y (y1, y2,$ etc.). Hence a common solution is obtained by assuming $y = c1y1 + c2y2+.....$ Now my questions are:

  1. How are linear differential equations formed from general algebraic expressions?
  2. Why is it that a linear differential equation of second or higher order contain more than one solution of y that satisfy them?
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If you're good with your linear algebra, solutions to an $n^{th}$ order linear differential equation form a vector space of dimension $n$, and every vector space has a basis of $n$ linearly independent vectors that span the space. In this case our vectors are functions! So an $n^{th}$ order equation will form an $n$-parameter family of solutions that are closed under addition and scalar multiplication. So if $y_1$,$y_2$,...,$y_n$ are solutions to the equation then $c_1y_1+c_2y_2+$...+$c_ny_n$ is also a solution for any real numbers $c_1,c_2$,...,$c_n$. This is also sometimes called the "principle of superposition".

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When we differentiate $f(x)$ and $f(x)+c$, for any constant$c$ we get the same derivative. If we are given that derivative we cannot say with definiteness for which functio9ns it is the derivative.That is the function $\cos x$ could be the derivative of either $\sin x$ or $\sin x +8$.

Now second order differential equation involves the second derivative: So the two functions $-\cos x +8x, -\cos x$ both when differentiated twice could give $\sin x$. These two are solutions of $y''= \cos x$ and are linearly independent.

To form a a linear differential equation from an algebraic expression is easy; The derivative of $\sin (x-a)^2$ is $2(x-a)\cos(x-a)^2$. Now we have to find an algebraic relationship between the function $y=\sin(x-a)^2$ and its derivative:

$2(x-a)(1-y^2)-y'= 0$

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