Help in solving a second-order linear differential equation

[Aero]

$$\frac{{d^2 y}}{{dx^2 }} + \left( {Ax + B} \right)y = 0$$

I have tried lots of substitions, but a solution won't pop out. Can anyone help solve this?

Thanks.

 

[Matthew Rodman]

Make the change of variable

$$Ax + B = \lambda u$$

(lambda is a constant) this will give you

$$\frac{A^2}{\lambda^2} \frac{d^2 y}{d u^2} + \lambda u y = 0$$

so if you then set

$$\lambda = -A^{\frac{2}{3}}$$

you then have

$$\frac{d^2 y}{d u^2} - u y = 0$$

which is the Airy equation (in u). Have a look on Wikipedia or elsewhere on Airy functions and such - or just type in "Airy Equation".

edit: note you will actually get three different solutions as the condition for lambda is

$$\frac{\lambda^3}{A^2} = -1$$

which means that there are three values of lambda that satisfy this (i.e. three distinct cube-roots) - one will be real-valued (already given) plus two complex ones.

Here is a link for the Airy function

http://mathworld.wolfram.com/AiryFunctions.html

 

[tacman]

Sure, I'd be happy to help with solving this second-order linear differential equation. The first step in solving any differential equation is to identify the type of equation and its order. In this case, we have a second-order linear differential equation, which means it has the form:

\frac{{d^2 y}}{{dx^2 }} + P(x)\frac{{dy}}{{dx}} + Q(x)y = R(x)

where P(x) and Q(x) are functions of x and R(x) is a function of y.

The next step is to check if the equation is homogeneous or non-homogeneous. A homogeneous equation has a R(x) term equal to 0, while a non-homogeneous equation has a non-zero R(x) term. In this case, our equation is homogeneous since R(x) = 0.

To solve a homogeneous second-order linear differential equation, we use the method of undetermined coefficients. This involves assuming a solution of the form y = e^{mx}, where m is a constant. We then substitute this into the equation and solve for m.

In our case, the equation becomes:

m^2e^{mx} + (Ax+B)e^{mx} = 0

We can simplify this to:

(m^2 + Am + B)e^{mx} = 0

Since e^{mx} is never equal to 0, the only way for this equation to hold is if the coefficient (m^2 + Am + B) is equal to 0. This gives us a quadratic equation in terms of m, which we can solve to find the values of m.

Once we have the values of m, we can then use them to find the general solution of the differential equation, which has the form:

y = c_1e^{m_1x} + c_2e^{m_2x}

where c_1 and c_2 are constants determined by any initial conditions given in the problem.

I hope this helps in solving your second-order linear differential equation. If you have any further questions or need clarification on any steps, please don't hesitate to ask. Good luck!