**Problem**: Use the substitution $ z = \sqrt{x} $ to solve the differential equation $ 4x \frac{d^2y}{dx^2}+2 \frac{dy}{dx}+y = 0 $.

**Solution**: Well, let's do what it says. From the chain rule, we have

$ \displaystyle \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = 2z \frac{dy}{dx} $.

Then we also have by the product rule and chain rule again

$ \displaystyle \frac{d^2y}{dz^2} = 2 \frac{dy}{dx}+2z \frac{d^2y}{dx^2} \cdot \frac{dx}{dz} = \frac{1}{z} \cdot \frac{dy}{dz}+4z^2 \frac{d^2y}{dx^2} $.

So we can make the substitutions

$ \displaystyle 4x \frac{d^2y}{dx^2} = 4z^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dz^2}-\frac{1}{z} \cdot \frac{dy}{dz} $

and

$ \displaystyle \frac{dy}{dx} = \frac{1}{2z} \cdot \frac{dy}{dz} $

to obtain the differential equation

$ \frac{d^2y}{dz^2}-\frac{1}{z} \cdot \frac{dy}{dz} + 2 \cdot \frac{1}{2z} \cdot \frac{dy}{dz}+y = \frac{d^2y}{dz^2}+y = 0 $.

But we know the solution to this is $ y = C_1 \cos{z}+C_2 \sin{z} $ so our final solution is

$ y = C_1 \cos{\sqrt{x}}+C_2 \sin{\sqrt{x}} $.

QED.

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Comment: This substitution is, of course, not really natural but was actually found after solving the ODE in another way. Fortunately, it simplifies the problem rather greatly and seems to be a useful technique to look out for.

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Practice Problem: Find another way to solve the differential equation.

Series. Lol.

ReplyDeleteanother way.. well there's the limit defination, and also the fundemental therome of calculus.

ReplyDeletebut there's the d^2y/dx^2, so i don't know if it works.

are any of these questions likely to appear on the SAT? or.. are there even any calculus on the SAT?

Nope; the SAT does not cover any material beyond Pre-Calculus and usually not even that unless you're taking the SAT Math Level 2.

ReplyDelete