Difference between revisions of "Advanced Methods to Solve 2nd-Order ODEs" - Rhea

Slightly we are moving to our last tutorial for the introduction of ordinary differential equations. This tutorial will give three particular methods to solve ODEs in the second order, which was also mentioned in 5.0. Theories are boring, we will be processing with sample questions.

8.1 Reduction of Order Method

This method is often used to find a second solution of an ODE, provided that one solution is given.

Example: Consier the ODE $ t^2 (lnt+1) \frac{d^2y}{dt^2} + t(2lnt+1) \frac{dy}{dt} -y=0 $, one of the solutions is $ y=\frac{1}{t} $. Find a second solution.

Solution: Try a general solution $ y=\frac{x}{t} $, where $ x=x(t) $, $ y=y(t) $.

Find the first and second derivative to $ y $, hence $ \frac{dy}{dt}=\frac{\frac{dx}{dt}}{t} - \frac{x}{t^2} $, $ \frac{d^2y}{dt^2}=\frac{2x}{t^3} - \frac{2\frac{dx}{dt}}{t^2} + \frac{\frac{d^2y}{dt^2}}{t} $.

Substitute them into ODE, hence $ t^2 (lnt+1) (\frac{2x}{t^3} - \frac{2\frac{dx}{dt}}{t^2} + \frac{\frac{d^2x}{dt^2}}{t}) + t (2lnt+1) (\frac{\frac{dx}{dt}}{t} - \frac{x}{t^2}) - \frac{x}{t} = 0 $,

Simplify it to have $ \frac{d^2x}{dt^2} t (lnt+1) - \frac{dx}{dt} = 0 $.

Let $ \frac{dx}{dt}=z $, then \frac{dz}{dt} (lnt+1) - \frac{z}{t} =0 </math>,

Solve the ODE in the first order by separating variables, hence $ \frac{dx}{dt} = z=A(lnt+1) $, where $ A $ is a non-zero real number.

Integrate with respect to $ t $, hence $ x=At lnt + B $, where $ A $, $ B $ are non-zero real numbers.

Plug it back to the general solution, $ y=\frac{x}{t} = A lnt + \frac{B}{t} $.

known that one of the solutions is $ y=\frac{1}{t} $, hence the second solution is $ y=lnt $.

8.2 Cauchy-Euler Equation

8.3 Varation of Constant

8.4 Exercises

8.5 References

Institute of Natural and Mathematical Science, Massey University. (2017). 160.204 Differential Equations I: Course materials. Auckland, New Zealand.