Question

Solve the differential equation `x^2(d^2y)/(dx^2) + x dy/dx + y = x^m`?

Answer 1

`y = Acos(lnx)+Bsin(lnx) + (x^m)/(m^2 + 1)`

Explanation:

We have:

`x^2(d^2y)/(dx^2) + x dy/dx + y = x^m` ..... [A]

This is a Euler-Cauchy Equation (the power of `x` is the same as the degree of the differential in every occurrence of their product) which is typically solved via a change of variable. Consider the substitution:

`x = e^t => xe^(-t)=1`

Then we have,

`dy/dx = e^(-t)dy/dt`, and, `(d^2y)/(dx^2)=((d^2y)/(dt^2)-dy/dt)e^(-2t)`

Substituting into the initial DE [A] we get:

`x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) + xe^(-t)dy/dt + y = (e^t)^m`

`:. ((d^2y)/(dt^2)-dy/dt) + dy/dt + y = e^(tm)`

`:. (d^2y)/(dt^2) + y = e^(tm)` ..... [B]

This is now a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complementary Function

`m^2+1 = 0`

We can solve this quadratic equation, and we get two complex conjugate solutions::

`m=+-i`

Thus the Homogeneous equation [B] has the solution:

`y_c = e^(0t)(Acos(1t)+Bsin(1t) )`
`\ \ \ =Acost+Bsint`

Particular Solution

With this particular equation [B], a probably solution is of the form:

`y = ae^(tm)`

Where `a` is a constant to be determined by substitution

Let us assume the above solution works, in which case be differentiating wrt `x` we have:

`y' \ \= ame^(tm)`
`y'' = am^2e^(tm)`

Substituting into the initial Differential Equation `[B]` we get:

`am^2e^(tm) + ae^(tm) = e^(tm)`
`:. am^2 + a = 1`
`:. a(m^2 + 1) = 1`
`:. a = 1/(m^2 + 1)`

And so we form the Particular solution:

`y_p = e^(tm)/(m^2 + 1)`

General Solution

Which then leads to the GS of [B}

`y(t) = y_c + y_p`
`\ \ \ \ \ \ \ = Acost+Bsint + e^(tm)/(m^2 + 1)`

Now we initially used a change of variable:

`x = e^t => t=lnx`

So restoring this change of variable we get:

`y = Acos(lnx)+Bsin(lnx) + ((e^t)^m)/(m^2 + 1)`
`\ \ = Acos(lnx)+Bsin(lnx) + ((x)^m)/(m^2 + 1)`

Which is the General Solution of [A].

`y = Acos(lnx)+Bsin(lnx) + (x^m)/(m^2 + 1)`