# Linearise the non linear d.e : h'=(qi-b*h^{1/2})/(\pi(2*r*h-h^2) about h(0) using taylor series...

## Question:

Linearise the non linear d.e :

{eq}h'=(qi-b*h^{1/2})/(\pi(2*r*h-h^2) {/eq}

about h(r=0) using taylor series expansion, just first two terms of taylor series.

## Linearise a Differntial Equation by Taylor Series:

The Taylor series of a (infinitely differentiable) function f (x) of a real variable x around the value {eq}x_0 {/eq}

{eq}f(x) = \sum_{n=0}^\infty \dfrac{1}{n!} f^{(n)}(x)\bigg|_{x=x_0} (x-x_0)^n {/eq}

allows a representation of the function as a power series of the variable in a neighborhood of the value {eq}x_0 {/eq}. Such representation can be truncated at any desired order of the expansion so as to provide an approximation easier to use in various applications.

In order to linearise the non-linear differential equation in the form

{eq}h'(r) = F( h(r) ) {/eq} ,

we construct the Taylor expansion of the right-hand side F ( h(r) ) around the value h ( 0 ) .

The first term of the expansion will be

{eq}F ( h (0) ) = \dfrac{ b \sqrt{ h (0) } - \text{qi}} { \pi h(0)^2} {/eq}

while the first order derivative evaluated at h(0) is

{eq}\dfrac{ \partial F(h) }{ \partial h} \bigg|_{h=h(0)} = \dfrac{4 \ h(0) \ \text{qi} - 3 \ b \ h(0)^{3/2}}{2 \ \pi \ h(0)^4} {/eq} .

The linearized differential equation is therefore

{eq}h'(r) = \dfrac{b \ \sqrt{ h(0) } - \text{qi}}{ \pi \ h(0)^2} + \left( \dfrac{4 \ h(0) \ \text{qi} - 3 \ b \ h(0)^{3/2} }{2 \ \pi \ h(0)^4} \right) h(r) {/eq}. 