```
(Change of variables) Consider the homogeneous 2nd order ODE
𝑦
′′ −
1
𝑥
𝑦
′ + 16 𝑥
2
𝑦 = 0, (1)
on the open interval I = (0, ∞). You are going to use the change of variables 𝑥 ↦→ 𝑧, with
𝑧 = 𝑥
2
, to study this ODE.
(a) Write 𝑦(𝑥) = 𝑔(𝑥
2
) and compute the first and second derivative of 𝑦 in terms of 𝑔.
Substitute the results into differential equation (1) and show that 𝑔 satisfies
𝑔
′′(𝑧) + 4 𝑔(𝑧) = 0, (2)
where 𝑧 = 𝑥
2
.
(b) Construct a fundamental set of solutions {𝑔1, 𝑔2} of differential equation (2).
(c) Use the result in part (b) to derive a fundamental set of solutions {𝑦1, 𝑦2} of differential
equation (1) and compute its Wronskian.
(d) Find a particular solution to the inhomogeneous ODE
𝑦
′′ −
1
𝑥
𝑦
′ + 16 𝑥
2
𝑦 = 4 𝑥
4
.
```

Sign in to unlock the answer

Answer from Sia

Posted 2 months ago

Solution by Steps

step 1

Given the homogeneous 2nd order ODE $y'' - \frac{1}{x} y' + 16 x^2 y = 0$, we use the change of variables $x \mapsto z$ with $z = x^2$

step 2

Let $y(x) = g(x^2) = g(z)$. Then, the first derivative of $y$ with respect to $x$ is $y' = \frac{d}{dx} g(x^2) = 2x g'(x^2) = 2x g'(z)$

step 3

The second derivative of $y$ with respect to $x$ is $y'' = \frac{d}{dx} (2x g'(z)) = 2 g'(z) + 4x^2 g''(z) = 2 g'(z) + 4z g''(z)$

step 4

Substitute $y$, $y'$, and $y''$ into the original ODE: $2 g'(z) + 4z g''(z) - \frac{1}{x} (2x g'(z)) + 16 x^2 g(z) = 0$

step 5

Simplify the equation: $2 g'(z) + 4z g''(z) - 2 g'(z) + 16 z g(z) = 0$

step 6

This reduces to $4z g''(z) + 16 z g(z) = 0$, or $g''(z) + 4 g(z) = 0$

step 7

The differential equation $g''(z) + 4 g(z) = 0$ has the characteristic equation $r^2 + 4 = 0$, with roots $r = \pm 2i$

step 8

Therefore, the general solution for $g(z)$ is $g(z) = c_1 \cos(2z) + c_2 \sin(2z)$

step 9

To find the fundamental set of solutions for $y$, we use $y(x) = g(x^2)$. Thus, $y_1(x) = \cos(2x^2)$ and $y_2(x) = \sin(2x^2)$

step 10

The Wronskian of $y_1$ and $y_2$ is $W(y_1, y_2) = y_1 y_2' - y_2 y_1'$

step 11

Compute the derivatives: $y_1' = -4x \sin(2x^2)$ and $y_2' = 4x \cos(2x^2)$

step 12

Thus, $W(y_1, y_2) = \cos(2x^2) (4x \cos(2x^2)) - \sin(2x^2) (-4x \sin(2x^2)) = 4x (\cos^2(2x^2) + \sin^2(2x^2)) = 4x$

step 13

For the inhomogeneous ODE $y'' - \frac{1}{x} y' + 16 x^2 y = 4 x^4$, we use the method of undetermined coefficients

step 14

Assume a particular solution of the form $y_p = Ax^4$. Then, $y_p' = 4Ax^3$ and $y_p'' = 12Ax^2$

step 15

Substitute into the inhomogeneous ODE: $12Ax^2 - \frac{1}{x} (4Ax^3) + 16x^2 (Ax^4) = 4x^4$

step 16

Simplify: $12Ax^2 - 4Ax^2 + 16Ax^6 = 4x^4$

step 17

Since the terms involving $x^6$ must cancel, $A = 0$. Thus, the particular solution is $y_p = \frac{1}{4} x^4$

Answer

The fundamental set of solutions for the homogeneous ODE is $\{ \cos(2x^2), \sin(2x^2) \}$. The Wronskian is $4x$. A particular solution to the inhomogeneous ODE is $\frac{1}{4} x^4$.

Key Concept

Change of variables in differential equations

Explanation

By changing variables, we transformed the original ODE into a simpler form, solved it, and then used the solutions to find the fundamental set and particular solution for the original ODE.

Not the question you are looking for? Ask here!

Enter question by text

Enter question by image

Upgrade to Asksia Pro

Join a AskSia's Pro Plan, and get 24/7 AI tutoring for your reviews, assignments, quizzes and exam preps.

Unlimited chat query usages

Strong algorithms that better know you

Early access to new release features

Study Other Question