I have the following third-order differential equation:- \begin{align*} &(x+1)\frac{d^3y}{dx^3} + 3\frac{d^2y}{dx^2}=0 \\ \\ & \text{such that, $y(1) = \frac{1}{2}$} \end{align*}
I have to show that the solution to the differential equation as series of ascending powers of $(x-1)$, up to and including the term in $(x-1)^3$ is, \begin{align*} y \approx \frac{1}{2} + \frac{3}{4}(x-1) + \frac{1}{8}(x-1)^2 - \frac{1}{16}(x-1)^3 \end{align*}
Can anyone help me solve this?
There are different methods to solve the ODE. I think the nicest way would be via order reduction:
Substitute $\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}^{2}x} = v\left( x \right)$ aka $\frac{\operatorname{d}^{3}y\left( x \right)}{\operatorname{d}^{3}x} = \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x}$:
$$ \begin{align*} \left( x + 1 \right) \cdot \frac{\operatorname{d}^{3}y\left( x \right)}{\operatorname{d}^{3}x} + 3 \cdot \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}^{2}x} &= 0\\ \left( x + 1 \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} + 3 \cdot v\left( x \right) &= 0\\ \end{align*} $$
With a bit of rearranging we would get this: $$ \begin{align*} \left( x + 1 \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} + 3 \cdot v\left( x \right) &= 0\\ \left( x + 1 \right) \cdot \frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x} &= -3 \cdot v\left( x \right)\\ \frac{\frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x}}{v\left( x \right)} &= -3 \cdot \frac{1}{x + 1}\\ \int \frac{\frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x}}{v\left( x \right)}\, \operatorname{d}x &= \int -3 \cdot \frac{1}{x + 1}\, \operatorname{d}x\\ \int \frac{\frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x}}{v\left( x \right)}\, \operatorname{d}x &= -3 \cdot \int \frac{1}{x + 1}\, \operatorname{d}x\\ \end{align*} $$
Now if we use $\int \left( \frac{\operatorname{d}^{n + 1}v\left( x \right)}{\operatorname{d}^{n + 1}x} \right) / \left( \frac{\operatorname{d}^{n}v\left( x \right)}{\operatorname{d}^{n}x} \right)\, \operatorname{d}x = \ln\left( \frac{\operatorname{d}^{n}v\left( x \right)}{\operatorname{d}^{n}x} \right)$ where $n = 0$ and $\int \frac{1}{a \cdot x + b}\, \operatorname{d}x = \frac{1}{a} \cdot \ln\left( a \cdot x + b \right)$ where $a = 1$ and $b = 1$ we get: $$ \begin{align*} \int \frac{\frac{\operatorname{d}v\left( x \right)}{\operatorname{d}x}}{v\left( x \right)}\, \operatorname{d}x &= \int -3 \cdot \frac{1}{x + 1}\, \operatorname{d}x\\ \ln\left( v\left( x \right) \right) &= -3 \cdot \ln\left( x + 1 \right) + c_{1}\\ v\left( x \right) &= \exp\left( -3 \cdot \ln\left( x + 1 \right) + c_{1} \right)\\ v\left( x \right) &= \left( \exp\left( \ln\left( x + 1 \right) \right) \right)^{-3} \cdot \exp\left( c_{1} \right)\\ v\left( x \right) &= \left( x + 1 \right)^{-3} \cdot \exp\left( c_{1} \right)\\ \end{align*} $$
Now Resubstitute $\frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}^{2}x} = v\left( x \right)$ and get: $$ \begin{align*} v\left( x \right) &= \left( x + 1 \right)^{-3} \cdot \exp\left( c_{1} \right)\\ \frac{\operatorname{d}^{2}y\left( x \right)}{\operatorname{d}^{2}x} &= \left( x + 1 \right)^{-3} \cdot \exp\left( c_{1} \right)\\ \frac{\operatorname{d}y\left( x \right)}{\operatorname{d}x} &= \int \left( x + 1 \right)^{-3} \cdot \exp\left( c_{1} \right)\, \operatorname{d}x\\ y\left( x \right) &= \int \int \left( x + 1 \right)^{-3} \cdot \exp\left( c_{1} \right)\, \operatorname{d}x\, \operatorname{d}x\\ y\left( x \right) &= \exp\left( c_{1} \right) \cdot \int \int \left( x + 1 \right)^{-3}\, \operatorname{d}x\, \operatorname{d}x\\ \end{align*} $$
Via using $\int f\left( a \cdot x + b \right)\, \operatorname{d}x = \frac{1}{a} \cdot F\left( a \cdot x + b \right)$ we get: $$ \begin{align*} y\left( x \right) &= \exp\left( c_{1} \right) \cdot \int \int \left( x + 1 \right)^{-3}\, \operatorname{d}x\, \operatorname{d}x\\ y\left( x \right) &= \exp\left( c_{1} \right) \cdot \int -\frac{1}{2} \cdot \left( x + 1 \right)^{-2} + c_{2}\, \operatorname{d}x\\ y\left( x \right) &= -\frac{\exp\left( c_{1} \right)}{2} \cdot \int \left( x + 1 \right)^{-2} + c_{2}\, \operatorname{d}x\\ y\left( x \right) &= -\frac{\exp\left( c_{1} \right)}{2} \cdot \left[ -\left( x + 1 \right)^{-1} + c_{2} \cdot x + c_{3} \right]\\ y\left( x \right) &= -\frac{\exp\left( c_{1} \right)}{2} \cdot \left[ -\frac{1}{x + 1} + c_{2} \cdot x + c_{3} \right]\\ y\left( x \right) &= \underbrace{\frac{\exp\left( c_{1} \right)}{2}}_{\text{constant}} \cdot \frac{1}{x + 1} + \underbrace{-\frac{\exp\left( c_{1} \right)}{2} \cdot c_{2}}_{\text{constant}} \cdot x + \underbrace{-\frac{\exp\left( c_{1} \right)}{2} \cdot c_{3}}_{\text{constant}}\\ y\left( x \right) &= c_{4} \cdot \frac{1}{x + 1} + c_{5} \cdot x + c_{6}\\ \end{align*} $$
$$\fbox{$y\left( x \right) = c_{4} \cdot \frac{1}{x + 1} + c_{5} \cdot x + c_{6}$}$$
With the condition $y\left( 1 \right) = \frac{1}{2}$ we get: $$ \begin{align*} y\left( 1 \right) &= c_{4} \cdot \frac{1}{1 + 1} + c_{5} \cdot 1 + c_{6}\\ \frac{1}{2} &= c_{4} \cdot \frac{1}{2} + c_{5} + c_{6}\\ \frac{1}{2} &= \frac{1}{2} \cdot c_{4} + c_{5} + c_{6}\\ \end{align*} $$