Question

Given a second order linear homogeneous differential equation a2(x)y′′+a1(x)y′+a0(x)y=0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y1,y2. But there are times when only one function, call it y1, is available and we would like to find a second linearly independent solution. We can find y2 using the method of reduction of order. First, under the necessary assumption the a2(x)≠0 we rewrite the equation as y′′+p(x)y′+q(x)y=0 p(x)=a1(x)a2(x), q(x)=a0(x)a2(x), Then the method of reduction of order gives a second linearly independent solution as y2(x)=Cy1u=Cy1(x)∫e−∫p(x)dxy21(x)dx where C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain y2=C3e2x then we can choose C=1/3 so that y2=e2x. Given the problem y′′−4y′+29y=0 and a solution y1=e2xsin(5x) Applying the reduction of order method we obtain the following y21(x)=Cy1u= e^(4x)sin^2(5x) p(x)= -4 and e−∫p(x)dx= x So we have ∫e−∫p(x)dxy21(x)dx=∫ 1 dx= x Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at y2(x)=Cy1u= So the general solution to y′′−5y′+4y=0 can be written as y=c1y1+c2y2=c1 +c2

Answer 1

Answer:

• eˆ(4*x/5)
• -20/25
• eˆ(0.8*x)
• eˆ(4*x/5)/[eˆ(4*x/5)]
• x
• x*eˆ(2*x/5)
• eˆ(0.4*x)
• x*eˆ(2*x/5)

Answer 2

The question is not clear. What is clear is that you are talking about solving differential equations using the method of reduction of order.

I will explain this method by solving the equation

y''- 4y' + 29 = 0

with y1 = e^(4x).

This would further help you to solve your problem if it is not in this question.

Step-by-step explanation:

Given the differential equation:

y''- 4y' + 29 = 0

with y1 = e^(4x)

To find the other solution using the method of reduction of order, we assume the second solution to be of the form

y2 = uy1 = ue^(4x)

Since this solution, just like the given solution, satisfies the given differential equation, then

y2'' - 4y2' + 29 = 0

y2' = u'e^(4x) + 4ue^(4x)

y2'' = u''e^(4x) + 4u'e^(4x) + 4u'e^(4x) + 16ue^(4x)

= u''e^(4x) + 8u'e^(4x) + 16ue^(4x)

y2'' - 4y2' + 29 = [u''e^(4x) + 8u'e^(4x) + 16ue^(4x)] - 4[u'e^(4x) + 4ue^(4x)] + 29

= u''e^(4x) + 4u'e^(4x) + 29 = 0

u'' + 4u' = -29e^(-4x)

Let w = u', then w' = u''

So

w' + 4w = -29e^(-4x)

Multiply both sides by the integrating factor e^(4x)

w'e^(4x) + 4we^(4x) = -29

d(we^(4x)) = -29

Integrating both sides

we^(4x) = -29x

w = -29xe^(4x)

But w = u'

u' = -29xe^(4x)

Integrating this, we have

u = (29/16)(1 - 4x)e^(4x) + C

Since y2 = uy1

The second solution is now

y2 = (29/16)(1 - 4x)e^(8x) + Ce^(4x)