Question

In: Advanced Math

Methods 1

Find two linearly independent solution of

y"+7xy=0. of the form

y1=1+a3x^3+a6x^6+....

y2=x+b4x^4+b7x^7+....

Enter the first few co-efficients

a3=

a6=

b4=

b7=

Expert Solution

Let

y1=1+a3x^3+a6x^6+...

y'1=3a3x^2+6a6x^5+...

y"1=6a3x+30a6x^4+...

So

7xy=7x+7a3x^4+7a6x^7+...

Then

y"+7xy=0

Hence

(6a3x+30a6x^4)+(7x+7a3x^4+7a6x^7)=0

(6a3x+7x)+(30a6x^4+7a3x^4)=0

6a3+7=0 and 30a6+7a3=0

If you evaluate we find that,

a3=-7/6 then a6=49/180

To find b4 and b7,let

y2=x+b4x^4+b7x^7+...

y'2=1+4bx^3+7b7x^6+...

y"2=12b4x^2+42b7x^5+...

So

7xy=7x^2+7b4x^5+7b7x^8+...

y"+7xy=0

Then

(12b4x^2+42b7x^5)+(7x^2+7b4x^5+7b7x^8)=0

12b4x^2+7x2=0 and

42b7x^5+7b4x^5=0

We evaluate to get

b4=-7/12

b7=7/72

a3=-7/6

a6=49/180

b4=-7/12

b7=7/72

Emmanuel Mahiga answered 2 years ago

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