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: Solve 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 $,

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

If an ODE is given in or can be rearranged to the form $ at^2 \frac{d^2y}{dx^2} + bt \frac{dy}{dx} +cy=f(t) $, where $ f(t) $ is a known function, it is a Cauchy-Euler equation and the below method can be applied.

Example: Solve the ODE $ 2t^2 \frac{d^2y}{dt^2} + 4t \frac{dy}{dt} + 5y = 17t^2 $.

Solution: First consider the homogeneous equation $ 2t^2 \frac{d^2y}{dt^2} + 4t \frac{dy}{dt} + 5y = 0 $.

Imitate the form of $ 17t^2 $ to try the solution $ y=t^k $, where $ k $ is a real number.

Find the first and second derivative, hence $ \frac{dy}{dt} = kt^{t-1} $, $ \frac{d^2y}{dt^2}= k(k-1)t^{k-2} $.

Substitute into the ODE, hence $ 2t^2 k(k-1) t^{k-2} + 4tk t^{k-1} + 5t^k = 0 $.

Simplify it to have $ t^k (2k^2 +2k +5)=0 $.

As $ t=0 $, $ y=0 $ is not a trivial solution, we have $ 2k^2 +2k +5=0 $.

Solve the algebraic equation to have $ k=- \frac{1}{2} ± \frac{3}{2}i $.

These are complex conjugate roots, hence we have the general solution for the homogeneous equation $ y_h = t^{-\frac{1}{2}} [A cos(\frac{3}{2} lnt) + B sin(\frac{3}{2} lnt)] $.

[Note: If we have distinct real roots for $ k $, then the general solution is $ y_h = At^{k_1}+Bt^{k_2}] $;

If we have repeated real roots for $ k $, then the general solution is $ y_h = t^k (A lnt +B) $;

If we have complex conjugate roots for $ k $, then the general solution is $ y_h =t^{\rho} [Acos(\omega lnt) + Bsin(\omega lnt)] $, where $ k=\rho ± i\omega $.]

Now try particular solution $ y_p = Ct^2 $ to the given ODE,

Find the first and second derivative, hence $ \frac{dy_p}{dt}=2Ct $, $ \frac{d^2y_p}{dt^2} = 2C $.

Substitute into the ODE to have

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.