Solve the following equations either by method (d)above or by assumingy=xk(or try both methods to compare them). See Problem 15.

(a)x2y''+3xy'-3y=0(b)x2y''+xy'-4y=0(c)x2y''+7xy'+9y=0(d)x2y''-xy'+6y=0

Short Answer

Expert verified

Insert proposed solution and find equation for k. Use results from problem 15 for special cases of k value.

Step by step solution

01

Given information

The given differential equation isx2y''+3xy'-3y=0

02

differential equation

A differential equation is an equation that relates one or more unknown functions and their derivatives.

03

solve for auxiliary equation

(a)

x2y''+3xy'-3y=0

Here, a2=1,a1=3and a0=-3. Therefore, kis given by:

k=-2±4+432

Thus k1=1andk2=-3. So a general solution for this problem has the form:

y=Ax+Bx-3

(b)

x2y''+xy'-4y=0

Here, a2=1,a1=1and a0=-4 Therefore, kis given by:k=0±0+442

Thus k1=2andk2=-2. So a general solution for this problem has the form:y=Ax2+Bx-2

(c)

x2y''+7xy'+9y=0

Here, a2=1,a1=7and a0=9. Therefore, kis given by:

k=-6±62+492

Thus k1=k2=-3. Here we have a case when both values of k are the same. From problem 15 we know that in such the second solution is given by xklnx. So the solution has the form:

y=x-3(A+Blnx)

(d)

x2y''-xy'+6y=0

Here, a2=1,a1=-1and a0=6. Therefore, k is given by:

k=2±22-462

Thus k1,2=1±i5. This time the values for k are complex and from problem 15 we know that the general solution is a trigonometric function (sine or cosine) in the form:

y1=xRe(k)cos(Im(k)x),y2=xRe(k)sin(Im(k)x)

Most general solution is a superposition of the latter solutions:

y=Axcos(5lnx)+Bxsin(5lnx)

Insert proposed solution and find equation for k . Use results from problem 15 for special cases of k value.

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