## Wednesday, October 25, 2006

### Serious Substitution. Topic: Calculus.

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.

#### 3 comments:

1. Series. Lol.

2. another way.. well there's the limit defination, and also the fundemental therome of calculus.
but 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?

3. 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.