Principle of superposition for a particular case of liner differential equation.

[arroy_0205]

Suppose a given second order linear differential equation has two different solutions; then it follows that their linear combination is also a solution. If y_1(x), y_2(x) are two solutions then y(x)=a_1y_1(x)+a_2y_2(x) is also a solution where a_1, a_2 are two constants.

If we consider the equation: \frac{d^2y(x)}{dx^2}+4\frac{dy(x)}{dx}+4y(x)=0, this has two equal solutions, y_{1}(x)=e^{-2x}=y_2(x) but the general solution is given by the nonlinear combination: y(x)=(a_1+a_2xe^{-2x}). This is not in accordance with the principle of superposition for linear equations. Can anybody please explain why this nonlinear combinations is being allowed in this case?

(Another issue, I have currently switched to ubuntu linux and using firefox. Why am I not able to see the equations properly? All I can see is the latex input commands without the desired output. What has gone wrong? This is happening with other posts containing math formula written in latex format in this forum.)

 

[LCKurtz]

Suppose a given second order linear differential equation has two different solutions; then it follows that their linear combination is also a solution.

Yes, for homogeneous equations.

If y_1(x), y_2(x) are two solutions then y(x)=a_1y_1(x)+a_2y_2(x) is also a solution where a_1, a_2 are two constants.

If we consider the equation: \frac{d^2y(x)}{dx^2}+4\frac{dy(x)}{dx}+4y(x)=0, this has two equal solutions, y_{1}(x)=e^{-2x}=y_2(x) but the general solution is given by the nonlinear combination: y(x)=(a_1+a_2xe^{-2x}).

xe^-2x is a solution, which is independent of your y₁ and y₂ (which aren't themselves independent).

This is not in accordance with the principle of superposition for linear equations. Can anybody please explain why this nonlinear combinations is being allowed in this case?

Yes it is. You have two solutions y₁ = y₂ and xe^-2x and you have a linear combination of them being a solution. The fact that xe^-2x itself is a nonlinear function is irrelevant.