# Differential equation solver

#### Vinod

Hello,
How to solve $4*x^2*y" +12*x*y' +3*y=0, y(4)=\frac18, y'(4)=\frac{-3}{64}$ with the help of Droid48 calculator application? It involves stiff initial value problem as well as vector-value differential equation.

#### Walagaster

MHF Helper
Hello,
How to solve $4*x^2*y" +12*x*y' +3*y=0, y(4)=\frac18, y'(4)=\frac{-3}{64}$ with the help of Droid48 calculator application? It involves stiff initial value problem as well as vector-value differential equation.
Try for a solution of the form $y=x^r$ and see what values of $r$ work. You shouldn't need a calculator for this.

2 people

#### SlipEternal

MHF Helper
The Wronskian is going to be $Ce^{-\displaystyle \int \dfrac{3}{x}dx} = Ce^{-3\ln x} = e^{\ln x^{-3k}} = x^{-3k}$ for some constant $k$. So, as Walagaster says, the solution will be $x^r$ for some $r$.

2 people

#### Vinod

The Wronskian is going to be $Ce^{-\displaystyle \int \dfrac{3}{x}dx} = Ce^{-3\ln x} = e^{\ln x^{-3k}} = x^{-3k}$ for some constant $k$. So, as Walagaster says, the solution will be $x^r$ for some $r$.
Hello,
How to compute wronskian? How it can be utilised to solve differential equations? Would you give me some more information of Wronskian. As far as i know it is a determinant that is related to linear independence of a set of differential equations.

#### Vinod

The Wronskian is going to be $Ce^{-\displaystyle \int \dfrac{3}{x}dx} = Ce^{-3\ln x} = e^{\ln x^{-3k}} = x^{-3k}$ for some constant $k$. So, as Walagaster says, the solution will be $x^r$ for some $r$.
Hello, So applying initial conditions $y=x^{-\frac32}$

#### Walagaster

MHF Helper
Hello, So applying initial conditions $y=x^{-\frac32}$
So, after 6 months and no . or ? after your quote, are you asking us or telling us? Do you have a question for us? Or are you hoping someone will check your work (which you didn't show) for you?

#### Archie

$$4x^2y''+12xy'+3y =0 \quad y(4)=\tfrac18, \, y'(4) = -\tfrac3{64}$$
\begin{align}
&\text{Let} & x &= e^t \\ &&\implies \tfrac{\mathrm dx}{\mathrm dt} &= e^t = x \\[8pt]
&\text{and} & y(t) &= y\big(x(t)\big) \\
&& \tfrac{\mathrm dy}{\mathrm dt} &= \tfrac{\mathrm dy}{\mathrm dx} \tfrac{\mathrm dx}{\mathrm dt} = x\tfrac{\mathrm dy}{\mathrm dx} &\implies \dot y &= xy' \\
&& \tfrac{\mathrm d^2y}{\mathrm dt^2} &= \tfrac{\mathrm dy}{\mathrm dx} \tfrac{\mathrm dx}{\mathrm dt} + x\tfrac{\mathrm d^2y}{\mathrm dx^2} \tfrac{\mathrm dx}{\mathrm dt} \\ &&&= x^2 \tfrac{\mathrm d^2y}{\mathrm dx^2} + x\tfrac{\mathrm dy}{\mathrm dx} &\implies \ddot y &= x^2y'' + xy' \end{align}
And so
\begin{align} 4x^2y''+12xy'+3y &= 4(x^2y''+ xy') + 8xy'+3y \\ &= 4\ddot y + 8\dot y + 3y = 0 \end{align}
The characteristic equation \begin{align} 4r^2 + 8r + 3 &= (2r)^2 + 4(2r) + 3 \\ &= (2r+1)(2r+3) = 0 \end{align}
Has roots $$r_1 = -\tfrac12 \quad \text{and} \quad r_2 = -\tfrac32$$
so \begin{align}y(t) &= c_1 e^{-\frac12 t} + c_2 e^{-\frac32 t} \\ &= c_1 (e^{t})^{-\frac12} + c_2 (e^t)^{-\frac32} \\ y(x) &= c_1 x^{-\frac12} + c_2x^{-\frac32 } \\ &= \frac{c_1}{\sqrt x} + \frac{c_2}{\sqrt{x^3}}\end{align}
And plug in the initial conditions to find the constants.

2 people

#### Vinod

$$4x^2y''+12xy'+3y =0 \quad y(4)=\tfrac18, \, y'(4) = -\tfrac3{64}$$
\begin{align}
&\text{Let} & x &= e^t \\ &&\implies \tfrac{\mathrm dx}{\mathrm dt} &= e^t = x \\[8pt]
&\text{and} & y(t) &= y\big(x(t)\big) \\
&& \tfrac{\mathrm dy}{\mathrm dt} &= \tfrac{\mathrm dy}{\mathrm dx} \tfrac{\mathrm dx}{\mathrm dt} = x\tfrac{\mathrm dy}{\mathrm dx} &\implies \dot y &= xy' \\
&& \tfrac{\mathrm d^2y}{\mathrm dt^2} &= \tfrac{\mathrm dy}{\mathrm dx} \tfrac{\mathrm dx}{\mathrm dt} + x\tfrac{\mathrm d^2y}{\mathrm dx^2} \tfrac{\mathrm dx}{\mathrm dt} \\ &&&= x^2 \tfrac{\mathrm d^2y}{\mathrm dx^2} + x\tfrac{\mathrm dy}{\mathrm dx} &\implies \ddot y &= x^2y'' + xy' \end{align}
And so
\begin{align} 4x^2y''+12xy'+3y &= 4(x^2y''+ xy') + 8xy'+3y \\ &= 4\ddot y + 8\dot y + 3y = 0 \end{align}
The characteristic equation \begin{align} 4r^2 + 8r + 3 &= (2r)^2 + 4(2r) + 3 \\ &= (2r+1)(2r+3) = 0 \end{align}
Has roots $$r_1 = -\tfrac12 \quad \text{and} \quad r_2 = -\tfrac32$$
so \begin{align}y(t) &= c_1 e^{-\frac12 t} + c_2 e^{-\frac32 t} \\ &= c_1 (e^{t})^{-\frac12} + c_2 (e^t)^{-\frac32} \\ y(x) &= c_1 x^{-\frac12} + c_2x^{-\frac32 } \\ &= \frac{c_1}{\sqrt x} + \frac{c_2}{\sqrt{x^3}}\end{align}
And plug in the initial conditions to find the constants.
Hello,
Your method of solving is different than mentioned in my PDF.

#### topsquark

Forum Staff
Hello,
Your method of solving is different than mentioned in my PDF.
Typically there is more than one way to solve a differential equation, though Archie's method is "standard." Can you show us what your pdf 's method looks like? Once we know that we can walk you through any difficulties you are having with it.

-Dan

1 person

#### Archie

Hello,
Your method of solving is different than mentioned in my PDF.
\begin{align}
&\text{Let} & t &= \ln x \\ &&\implies \tfrac{\mathrm dt}{\mathrm dx}&= \tfrac1x \\[8pt]
&\text{and} & y(x) &= y\big(t(x)\big) \\
&& \tfrac{\mathrm dy}{\mathrm dx} &= \tfrac{\mathrm dy}{\mathrm dt} \tfrac{\mathrm dt}{\mathrm dx} = \tfrac1x\dot y &\implies xy' &= \dot y \\
&& \tfrac{\mathrm d^2y}{\mathrm dx^2} &= -\tfrac1{x^2}\tfrac{\mathrm dy}{\mathrm dt} + \tfrac1{x}(\tfrac{\mathrm d^2y}{\mathrm dt^2}\tfrac{\mathrm dt}{\mathrm dx}) \\ &&&= \tfrac1{x^2}\ddot y - \tfrac1{x^2}\dot y &\implies
x^2y'' &= \ddot y - \dot y\end{align}
And so
\begin{align} 4x^2y''+12xy'+3y &= 4(\ddot y - \dot y) + 12\dot y + 3y \\ &= 4\ddot y + 8\dot y + 3y = 0 \end{align}
And continue as before.

2 people