Question:

Use the substitution \(y=z/x\) to transform the differential equation
 \[x\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+(2-4x)\frac{\mathrm{d}y}{\mathrm{d}x}-4y=0\] into the equation \[\frac{\mathrm{d}^2 z}{\mathrm{d}x^2}-4\frac{\mathrm{d}z}{\mathrm{d}x}=0\]

Hence solve the equation \(x\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}+(2-4x)\frac{\mathrm{d}y}{\mathrm{d}x}-4y=0\), giving \(y\) in terms of \(x\).

Answer:

For the sake of ease, I’ll rewrite the derivative of \(y\) and \(z\) with respect to \(x\) as \(y’\) and \(z’\) respectively.

So first, take the first and second derivatives of our substitution with respect to \(x\).

\[y=\frac{z}{x}\]

\[y’=\frac{z’}{x}-\frac{z}{x^2}\]

\[y”=\frac{z”}{x}-\frac{2z’}{x^2}+\frac{2z}{x^3}\]

The trick there was to remember that \(z\) is not a constant, it is a function of \(x\) so you have to use the product (or quotient) rule to differentiate it.

Now substitute it into our original equation

\[xy”+(2-4x)y’-4y=0\]

\[x\left(\frac{z”}{x}-\frac{2z’}{x^2}+\frac{2z}{x^3}\right)+(2-4x)\left(\frac{z’}{x}-\frac{z}{x^2}\right)-\frac{4z}{x}=0\]

\[z”-\frac{2z’}{x}+\frac{2z}{x^2}+\frac{2z’}{x}-\frac{2z}{x^2}-4z’+\frac{4z}{x}-\frac{4z}{x}=0\]

\[z”-4z’=0\]

Now one can solve this like any 2nd order ODE to get \(z=A+Be^{4x}\) where \(A\) and \(B\) are some constants. Now we can arrange our original substitution and stick it back in to get \[xy=A+Be^{4x}\]

\[y=\frac{A+Be^{4x}}{x}\]

Which is the correct answer. I hope that helps and don’t hesitate to ask any further questions.

forwardslashreality:

JOKE #5 

A: “What is the integral of 1/cabin?”

B: “log cabin.”

A: “Nope, houseboat—you forgot the C.”

So bad, yet so good.

I love the one about chickens crossing the road.

dieplzkthxbye:

ok I know this is simple, but I cant quite remember how to go about it and I refuse to just guess, I want to do it the right way.

inverse f^(-1)(x)  of f(x)=2+ sqrt (6x+5)

I think stevanous has made a mistake in his calculation.

To restate the question, find the inverse \(f^{-1}(x)\) of \(f(x) = 2 + \sqrt{6x+5}\).

The easiest way to do this is let \(y = 2 + \sqrt{6x+5}\) and rearrange for \(x\). So first things first:

\[y - 2 = \sqrt{6x+5}\]

\[(y-2)^2 = 6x + 5\]

\[6x = (y-2)^2 - 5\]

\[x = \frac{(y-2)^2 - 5}{6}\]

\[x = \frac{1}{6}(y^2 - 4y - 1)\]

Now replace \(x\) with \(y\) and vise versa to get:

\[f^{-1}(x) = y = \frac{1}{6}(x^2 - 4x - 1)\]

Checking our answer:

\[f\left(f^{-1}(x)\right) = 2 + \sqrt{6\left(\frac{1}{6}(x^2 - 4x - 1)\right)+5}\]

\[f\left(f^{-1}(x)\right) = 2 + \sqrt{(x^2 - 4x - 1)+5}\]

\[f\left(f^{-1}(x)\right) = 2 + \sqrt{(x-2)^2}\]

\[f\left(f^{-1}(x)\right) = 2 + x - 2 = x\]

So it is the inverse, I hope that helped.

meerabai:

You won’t believe that these are all true.

I proved some of these to be true in my spring term class-but they’re explained really well in this little slideslow. Check it out if you’re interested :) 

aguide2fantasy:

Okay there is a demon (100% demon), and he had a child with a human (100%) human. Their daughter is 50% demon. So far so good.

The daughter (50% demon) has a child with a human (100% human). That child is 25% demon. All good.

That child (25% demon) has a daughter with a full demon (100% demon). What % of the daughter is demon? 

(I’m a writer and it is important to the story and i need an answer and I’ve spent like 2 hours trying to figure this out)

The answer is 62.5% demon and 37.5% human. Also, there is a quick and easy way of working out what I’m going to call heritage problems (I don’t know if they have a proper name). More after the cut.

Read More

kanrose:

If you can pronounce correctly every word in this poem, you will be speaking English better than 90% of the native English speakers in the world.

After trying the verses, a Frenchman said he’d prefer six months of hard labour to reading six lines aloud.

[source]

consulting-dick:

H e l p
I’m supposed to use sin^2x+cos^2x= 1
Aaaaahahdbd

Actually yes, that’s exactly what you have to do. But first rearrange it into the form:

\[\cos^{2}\theta = 1 - \sin^{2}\theta\]

Substitute into our original equation to get:

\[3(1 - \sin^{2}\theta) = \sin\theta + 1\]

Now expanding and rearranging gives:

\[3 - 3\sin^{2}\theta = \sin\theta + 1\]

\[3 = 3\sin^{2}\theta + \sin\theta + 1\]

\[3\sin^{2}\theta + \sin\theta - 2 =0\]

The second part follows from the first (Hint: let \(x = \sin\theta\) and see if you recognise the resulting equation).

Hope that helps

So I was recently asked a question about angles and triangles which goes as follows: Given the triangle below

Where \(x\) is unknown. Prove:

\[s\times \cos (a) = \sqrt{s^2 - h^2}\]

Firstly, lets write down \(\cos(\theta)\) for any angle \(\theta\).

\[\cos(\theta) = \frac{\text{Adjacent Side}}{\text{Hypotenuse}}\]

So it follows that:

\[\cos(a) = \frac{x}{s}\]

But we don’t know \(x\). So we try Pythagoras’ Theorem (\(s^2 = x^2 + h^2\)) to get that:

\[x^2 = s^2 - h^2\]

\[\Rightarrow x = \sqrt{s^2 - h^2}\]

Now we can write \(\cos(a)\) as:

\[\cos(a) = \frac{\sqrt{s^2 - h^2}}{s}\]

\[\Rightarrow s\times \cos(a) = \sqrt{s^2 - h^2}\]

So it is shown.

wtf-fun-factss:

If you cut a rose stem 4x vertically and soak each part in a different colored dye you get a rainbow rose thanks to the Fibonacci sequence. 

novellius:

Why do we British say Maths and Americans say Math? It seems so random help

Well, the word orignally comes from the greek máthēma but came to English via the Latin pural mathematica meaining roughly ‘all things mathematical’. So the history of our word mathematics is a plural. But we now use it as a singular because English wouldn’t be English without linguistic bastardisations.

So when we shorten it in British English, we keep the s. Some argue that it is because there are multiple disciplines in mathematics (Geometry, Calculus, and Algebra being some of the heavy hitters) so we should keep the plural. Personally I prefer the softer sound of the s but what can you do.

In short it is just another divergence of British English and American English which is somewhat inevitable I guess.

(Although, Americans do seem to have an affinity for changing ‘s’ to ‘z’ and ‘ph’ to ‘f’ for some reason unbeknownst to me)

managerie76:

madlori:

I bet you did.

Did you estimate what mpg your car is getting?  Did you figure out how many hamburgers you could get for six bucks?  Did you think about how long it would take you to get somewhere given the speed and distance?  Did you plan a meal so all the components would be done at the same time?  Did you encounter anything on sale?  Did you figure out how many groceries you could get for how much money you had?

There are about a million other daily, unconscious tasks that use algebra or at least algebraic thinking.  Just because you weren’t writing out an equation or employing variables doesn’t mean you weren’t using the skills that algebra and other math courses taught you.

Science and math aren’t important because you’re going to need to know the exact steps of photosynthesis or the quadratic formula.  They’re important because they teach you scientific and mathematical literacy and rational thinking, and that is sorely needed in a world where charlatans and cheats or people with a political or religious agenda can get away with all manner of pseudoscience and bullshit because people don’t have enough scientific literacy or critical thinking skills to accurately weigh the arguments or even discern where they fail logically.

So study math and science, and art, and literature, and history, and politics, not because you’re going to need it or it’s going to do something specific for you, but because an uninformed populace is bad for the world.