Question

Differential Equations help?? I don't see how using substitution x=lnt t u(x)=y(e^x) into t2 y'' + t y' + y = 0 we can get u''+u=0

Answer 1

I was given x=lnt I don't know how they came up with u(x)=y(e^x) or what that means

Answer 2

\[(t^2\ y'' + t\ y' + y = 0)/t^2\]
\[y'' + t^{-1}\ y' +t^{-2} y = 0\]
hmmm, whats the name of the method you think that used?

Answer 3

.... think that they used. cant type

Answer 4

if i were to take a guess at this, since I aint to adept at these:

\[r^2+t^{-1}r+t^{-2}=0\]

\[r=\frac{-t^{-1}\pm \sqrt{t^{-2}-4(t^{-2})}}{2}\]
\[r=\frac{-t^{-1}\pm \sqrt{t^{-2}(1-4)}}{2}\]
\[r=\frac{-t^{-1}\pm t^{-1}\sqrt{-3}}{2}\]
that cant be good ....

Answer 5

With substituting \(t=\ln{x}\), it follows that
\(\large{\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=\frac{1}{t}\frac{dy}{dx}}\), and
\(\large{\frac{d^2y}{dt^2}=\frac{1}{t^2}\frac{d^2y}{dx^2}-\frac{1}{t^2}\frac{dy}{dx}=\frac{1}{t^2}(\frac{d^2y}{dx^2}-\frac{dy}{dx})}\) (By using product rule).
Substitute that into the differential equation.

Answer 6

It looked like this when I was using chrome, but with firefox it's looking good!

Answer 7

Let me write it again. All I did was to use the substitution \(x=\ln t\), to find the derivatives with respect to x.

Answer 8

\[{dy \over dt}={1 \over t}{dy \over dx}, \text { and } {d^2y \over dt^2}={1 \over t^2}({d^2y \over dx^2}-{dy \over dx})\].

Answer 9

If we substitute this into the differential equation, we get:
\[{d^2y \over dx^2}-{dy \over dx}+{dy \over dx}+y=0 \implies {d^2y \over dx^2}+y=0.\]
Which is the same as the equation you have.

Answer 10

All you need now is to use the auxiliary equation to find the general solution, and then substitute back for \(t\).

Answer 11

Sorry I'm still new at this I'm lost when you did the d^2y/dt^2=1/t^2(d^2y/dx^2−dydx)

Answer 12

And when you plug into the differential equation t2 y'' + t y' + y = 0 where did the t go?

Answer 13

Just use the product rule to find \(\large{\frac{d^2y}{dt^2}}\).

Answer 14

The t's were cancelled out with the t's that are in the denominators of \(\large{\frac{d^2y}{dt^2}}\) and \(\large {\frac{dy}{dt}}\).

Answer 15

but for d^2y/dt^2 shouldn't it be \[d^2y/dt^2=1/t^2(d^2y/dx^2 - 1/t^2 dy/dx)\]

Answer 16

\[d^2y/dt^2=1/t^2(d^2y/dx^2−1/t dy/dx)\]

Answer 17

sorry I meant the last one I posted shouldn't there still be a 1/t in the brackets