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kevin30024's Question
Math
Posted 2 months ago
(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
.
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Answer from Sia
Posted 2 months ago
Solution by Steps
step 1
Given the homogeneous 2nd order ODE y′′−1xy′+16x2y=0 y'' - \frac{1}{x} y' + 16 x^2 y = 0 , we use the change of variables x↦z x \mapsto z with z=x2 z = x^2
step 2
Let y(x)=g(x2)=g(z) y(x) = g(x^2) = g(z) . Then, the first derivative of y y with respect to x x is y′=ddxg(x2)=2xg′(x2)=2xg′(z) y' = \frac{d}{dx} g(x^2) = 2x g'(x^2) = 2x g'(z)
step 3
The second derivative of y y with respect to x x is y′′=ddx(2xg′(z))=2g′(z)+4x2g′′(z)=2g′(z)+4zg′′(z) 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 , y′ y' , and y′′ y'' into the original ODE: 2g′(z)+4zg′′(z)−1x(2xg′(z))+16x2g(z)=0 2 g'(z) + 4z g''(z) - \frac{1}{x} (2x g'(z)) + 16 x^2 g(z) = 0
step 5
Simplify the equation: 2g′(z)+4zg′′(z)−2g′(z)+16zg(z)=0 2 g'(z) + 4z g''(z) - 2 g'(z) + 16 z g(z) = 0
step 6
This reduces to 4zg′′(z)+16zg(z)=0 4z g''(z) + 16 z g(z) = 0 , or g′′(z)+4g(z)=0 g''(z) + 4 g(z) = 0
step 7
The differential equation g′′(z)+4g(z)=0 g''(z) + 4 g(z) = 0 has the characteristic equation r2+4=0 r^2 + 4 = 0 , with roots r=±2i r = \pm 2i
step 8
Therefore, the general solution for g(z) g(z) is g(z)=c1cos⁡(2z)+c2sin⁡(2z) g(z) = c_1 \cos(2z) + c_2 \sin(2z)
step 9
To find the fundamental set of solutions for y y , we use y(x)=g(x2) y(x) = g(x^2) . Thus, y1(x)=cos⁡(2x2) y_1(x) = \cos(2x^2) and y2(x)=sin⁡(2x2) y_2(x) = \sin(2x^2)
step 10
The Wronskian of y1 y_1 and y2 y_2 is W(y1,y2)=y1y2′−y2y1′ W(y_1, y_2) = y_1 y_2' - y_2 y_1'
step 11
Compute the derivatives: y1′=−4xsin⁡(2x2) y_1' = -4x \sin(2x^2) and y2′=4xcos⁡(2x2) y_2' = 4x \cos(2x^2)
step 12
Thus, W(y1,y2)=cos⁡(2x2)(4xcos⁡(2x2))−sin⁡(2x2)(−4xsin⁡(2x2))=4x(cos⁡2(2x2)+sin⁡2(2x2))=4x 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′′−1xy′+16x2y=4x4 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 yp=Ax4 y_p = Ax^4 . Then, yp′=4Ax3 y_p' = 4Ax^3 and yp′′=12Ax2 y_p'' = 12Ax^2
step 15
Substitute into the inhomogeneous ODE: 12Ax2−1x(4Ax3)+16x2(Ax4)=4x4 12Ax^2 - \frac{1}{x} (4Ax^3) + 16x^2 (Ax^4) = 4x^4
step 16
Simplify: 12Ax2−4Ax2+16Ax6=4x4 12Ax^2 - 4Ax^2 + 16Ax^6 = 4x^4
step 17
Since the terms involving x6 x^6 must cancel, A=0 A = 0 . Thus, the particular solution is yp=14x4 y_p = \frac{1}{4} x^4
Answer
The fundamental set of solutions for the homogeneous ODE is {cos⁡(2x2),sin⁡(2x2)} \{ \cos(2x^2), \sin(2x^2) \} . The Wronskian is 4x 4x . A particular solution to the inhomogeneous ODE is 14x4 \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.

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