Question

solve the differential equation: d^2y/dx^2 - dy/dx + e^(2x)y = 0

Answer 1

Substitute as usual an Ansatz/trial solution of \( y(x) = e^{rx} \) and then solve for \( r \). As this equation is second-order and linear, there will be two linearly independent solutions.

Answer 2

i can't solve it .. tried what you said .. =\

Answer 3

Sorry, I made a mistake. That won't work.

Instead, you're going to need to figure out how to get rid of that coefficient of y. So instead, let's try a substitution: \( u = e^x \). Then the equation is
y'' - y' + u^2y = 0

Now...

Answer 4

you need to express the equation in terms of derivatives of u, not x. For example,
\[ \frac{dy}{dx} = \frac{dy]{du}\frac{du}{dx} \]
and you know that du/dx = u. Hence
\[ y' = \frac{dy}{dx} = u \frac{dy}{du} \]

Calculate also the second derivative from this formula and then substitute it into the original equation and simplify.

Answer 5

hmm . i have a question about it .. if i do u=e^x then i can't say that y is only a function of u right ? if i will say dy/dx= dy/dt*dt/dx i will loose maybe the differentials of other x factors no ?

Answer 6

\[y''-y'+e^{2x}y=0\]as James said let\[u=e^x\]then we have (using the chain rule)\[y'=\frac{dy}{du}\cdot\frac{du}{dx}=u\frac{dy}{du}\]and\[y''=\frac{d^2y}{du^2}\cdot(\frac{du}{dx})^2+\frac{dy}{du}\cdot\frac{du}{dx}=u^2\frac{d^2y}{du^2}+u\frac{dy}{du}\]subbing in to our equation we get\[u^2\frac{d^2y}{du^2}+u\frac{dy}{du}-u\frac{dy}{du}+u^2y=0\]\[u^2\frac{d^2y}{du^2}+u^2y=0\]\[\frac{d^2y}{du^2}+y=0\]the associated polynomial is\[r^2+1=0\to r=\pm i\]the form of the solution will be\[y=c_1\cos u+c_2\sin u\]subbing back in terms of x we have\[y=c_1\cos(e^x)+c_2\sin(e^x)\]

Answer 7

can you please explain more on how you did the y'' part ? =]

Answer 8

We have to take the derivative w.r.t. x of both sides of the equation
\[ y' = u. dy/du \]
Hence using the product rule,
\[ y'' = \frac{du}{dx} \frac{dy}{du} + u \frac{d \ }{dx} \left( \frac{dy}{du} \right)\]
Now we already know from \( u = e^x \) that \( du/dx = u \); that's the first term taken care of.

For the second term, we have to use the chain rule again:
\[ \frac{d \ }{dx} \left( \frac{dy}{du} \right) = \frac{du}{dx} \frac{d \ }{du} \left( \frac{dy}{du} \right) = u \frac{d^2y}{du^2} \]
Now reassemble the pieces.