Find constants from given particular solution of a DE

Question

Give DE: x²y'' + bxy' + cy=0, find b and c such that 5x⁷ln(x) is a particular solution

b) Given x²y'' + bxy' + cy= 6x¹¹ for the values of b and c from part a), find a particular solution

Answer 1

a)

Call the given solution

f(x) = 5x⁷ln(x)

First calculate

f'(x) = 5(7x⁶ln(x) + x⁷/x) = 5x⁶(7ln(x) + 1)

f"(x) = 30x⁵(7ln(x) + 1) + 5x⁶ 7/x = 5x⁵(42ln(x) + 13)

We will plug into the DE y = f(x), which will give us constraints on b and c.

5x⁷(42ln(x) + 13) + 5bx⁷(7ln(x) + 1) + 5cx⁷ln(x) = 0

For this identity to hold for all x, we can divide by 5x⁷:

42ln(x) + 13 + b7ln(x) + b + cln(x) = 0

(42 + 7b + c)ln(x) + 13 + b = 0

For this identity to hold for all x, both coefficients must be 0

42 + 7b + c = 0

13 + b = 0

The solution is

b = -13

c = 49

b)

I do not know whether you were given some systematic way of solving the modified equation,

but it is easy to guess that the solution will be of the form

g(x) = f(x) + dx¹¹

for some d.

We just need to plug y = g(x) into the modified DE to calculate d.

First calculate

g'(x) = f'(x) + 11dx¹⁰

g"(x) = f"(x) + 110dx⁹

Plug g(x) in:

x²f"(x) + 110dx¹¹ + bxf'(x) + 11bdx¹¹ + cf(x) + cdx¹¹ = 6x¹¹

Use the fact that f(x) satisfies the original DE

110dx¹¹ + 11bdx¹¹ + cdx¹¹ = 6x¹¹

d = 6/(110 + 11b + c)

= 6/(110 - 143 + 49) = 0.375

Answer 2

Hi Ramesh,

I will be answering part 1 of this.

You're given a particular solution y(x) = 5x⁷ln(x) so you will be plugging it in to the diff. eq.

First, you need to find the first and 2nd derivatives, which will require you to apply the sum rule and product rule for derivatives:

y = 5x⁷ln(x)

y'= 35x⁶*lnx + (5x⁷/x)= 35x⁶*lnx + 5x⁶

y" = 210x⁵*lnx + 35x⁶/x+30x⁵ = 210x⁵*lnx + 35x⁵+30x⁵ = 210x⁵*lnx + 65x⁵

Now to plug it in the DE, you would need x2*y", bxy' and cy:

x²y"=x²(210x⁵*lnx + 65x⁵)= 210x⁷*lnx + 65x⁷

bxy'= bx(35x⁶*lnx + 5x⁶)=35bx⁷*lnx + 5bx⁷

cy=5cx⁷ln(x)

Now to add them all in the DE you were given: x²y'' + bxy' + cy=0

Meaning: 210x⁷*lnx + 65x⁷ + 35bx⁷*lnx + 5bx⁷ + 5cx⁷ln(x) =0

Adding like terms, you get: [210x⁷*lnx +35bx⁷*lnx + 5cx⁷ln(x)]+ [65x⁷ + 5bx⁷ ] =0

Let's start with the last portion, as it looks the easier one of the two:

factoring out the x7, you get x⁷(65+5b) = 0 so 65=-5b so b = -13

Now with the other portion, we can identify common factors and replace b by -13:

x⁷*lnx(210+35b+5c)=0 so 210 + 35b +5c =0

b = -13 so 210 + 35(-13) + 5c = 0 so 210-455+5c = 0 so -245 + 5c = 0 so 5c = 245 so c = 49

To recap part 1: b = -13 and c = 49

Cheers!