Question

the normal form of the differential equation y"-xy'+(4x^2-1)y=?

Answer 1

Answer on Question #70612, Math / Differential Equations

The normal form of the differential equation

$$y'' - x y' + (4 x^2 - 1) y = 0$$

Solution

Consider a second order linear ODE in the standard form

$$y'' + p(x) y' + q(x) y = 0$$

By a change of dependent variable, (1) can be written as

$$u'' + Q(x) u = 0$$

which is called the normal form of (1).

To find the transformation, let us put $y(x) = u(x)v(x)$.

When this is substituted in (1), we get

$$y' = u' v + v' u$$ $$y'' = u'' v + u' v' + v'' u + v' u'$$ $$v u'' + (2 v' + p v) u' + (v'' + p v' + q v) u = 0$$

Now we set the coefficient of $u'$ to zero. This gives

$$2 v' + p v = 0 \Rightarrow \frac{d v}{v} = -\frac{p}{2} d x \Rightarrow v = e^{-\int p / 2 d x}$$

Now coefficient of $u$ becomes

$$\left(q - \frac{1}{4} p^2 - \frac{1}{2} p'\right) v = Q(x) v$$

Since $v$ is nonzero, cancelling $v$ we get the required normal form. Also, since $v$ never vanishes, $u$ vanishes if and only if $y$ vanishes. Thus, the above transformation has no effect on the zeros of solution.

$$y'' - x y' + (4 x^2 - 1) y = 0$$

We have that

$$p(x) = -x, q(x) = 4 x^2 - 1$$

Then

$$v = e^{-\int (-x)/2 d x} = e^{x^2 / 4}$$

Now

$$Q(x) = 4 x^2 - 1 - \frac{1}{4} (-x)^2 - \frac{1}{2} (-x)''$$ $$Q(x) = 4 x^2 - 1 - \frac{1}{4} x^2 + \frac{1}{2}$$ $$Q(x) = \frac{15}{4} x^2 - \frac{1}{2}$$

Thus, the equation in normal form becomes

$$u'' + \left(\frac{15}{4} x^2 - \frac{1}{2}\right) u = 0,$$

where $u = y e^{-x^2 / 4}$

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