Explain 6-12 .11 T-Mobile 1:18 PM @ 70% Week #6 Test 2 Review sin c or is presen

Question

Explain 6-12 .11 T-Mobile 1:18 PM @ 70% Week #6 Test 2 Review sin c or is present in the differential equation? 6. What does the Wronskian tell us about our solutions? What should we do with this information? 7. When should we use Variation of Parameters? When should we use tables or "guesses"? 8. Why should we add another x to our particular solution guess for repeated roots? 9. Explain why we solve the characteristic equation in order to find the homogeneous solution. Why and how does it work? 10. What is the reason for us switching c1 and c2 with v1 and v2 for the particular equation? 11. Write the three generalized solution forms for 2nd order homogeneous differential equations 12. A particular solution that simplifies to zero is not possible? T/F Explain Previous Next Courses Calendar To Do Notifications Inbox

Explanation / Answer

If the functions fi are linearly dependent, then so are the columns of the Wronskian as differentiation is a linear operation, so the Wronskian vanishes. Thus, the Wronskian can be used to show that a set of differentiable functions is linearly independent on an interval by showing that it does not vanish identically. It may, however, vanish at isolated points.

A common misconception is that W = 0 everywhere implies linear dependence, but Peano (1889) pointed out that the functions x^2 and |x|x have continuous derivatives and their Wronskian vanishes everywhere, yet they are not linearly dependent in any neighborhood of 0.

In general, for an nth order linear differential equation, if ( n 1 ) solutions are known, the last one can be determined by using the Wronskian.

Consider the second order differential equation in Lagrange's notation

    y = a y + b y

where a(x),b(x) are known functions of x and y(x) is the yet to be determined function. Let us call y 1 , y 2 the two solutions of the equation and form their Wronskian

    W ( x ) = y 1 y 2 y 2 y 1

Then differentiating W(x) and using the fact that y i obey the above differential equation shows that

    W ( x ) = a W ( x )

Therefore, the Wronskian obeys a simple first order differential equation and can be exactly solved:

    W ( x ) = e^{A(x)}

where A'(x)=a(x)

Now suppose that we know one of the solutions y2 . Then, by the definition of the Wronskian, y 1 obeys a first order differential equation:

    y 1 (y 2 / y 2) y 1 = W ( x ) / y 2

and can be solved exactly

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