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

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)

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it