Question

Find the general solutions of the differential equations in problems.

y"-4y=0

Answer 1

We have not discussed this material yet in class just want to know how to solve these equations.

Answer 2

@ganeshie8

Answer 3

Hello! Glad to see you back here on OpenStudy.
\[y"-4y=0\] is a second order differential equation (since the highest derivative is the second).

I'll give you a quick overview of the basics of solving such a d. e., necessarily leaving out details that you'll soon pick up in class or from your text or online study materials.

We write an "auxiliary equation," as follows:

For the d. e. \[y"-4y=0 \rightarrow 1y" + 0y' - 4y,\] we borrow the coefficients 1, 0 and -4, and replace y" with m^2, y' with m and y with1, where k is a constant. The result here is 1m^2 + 0m - 4 = 0, all in m. Like any other quadratic equation, the roots of this one could be real, imaginary or complex.

Our auxiliary equation here is m^2-4=0, which has two real, unequal roots, m=2 and m=-2.

The general solution to the given d. e. is \[y=C _{1}e ^{m _{1}x}+C _{2}e ^{m _{2}x}.\]In our specific example, with the m values being plus or minus 2, the general solustion is\[y=C _{1}e ^{2x}+C _{2}e ^{-2x}.\] Unless we have to worry about "initial conditions," that's it.

Once again, the roots of the aux. eq'n. could be real and unequal (as ours are in this first example), real and equal, imaginary, or complex. Soon we'll go over the solutions that result in each case.

Answer 4

Ask all the questions you like, now or after this approach to solving 2nd order differential equations has been discussed in class.

Answer 5

It looks a lot like the problems we were solving earlier. Instead we were replacing y" with r^2*e^(rx), y' with r*e^(rx), and y with e^(rx).