solve: dy/dx+4xy=e^-2x & y(0)=1

Question

dy/dx+4xy=e^-2x & y(0)=1

Answer 1

Let Y,h: Y'h+4xy,h=0

Answer 2

→ {dY,h/Y,h={-4xdx

Answer 3

→lny,h=-2x^2+c ~upper

Answer 4

Y,h=c•e^-2x^2

Answer 5

Now if F_=F(x):

Answer 6

Y=F•e^-2x^2 ,

Answer 7

Y'=(F'-4xF)e^-2x^2

Answer 8

And by substitution to the given:

Answer 9

F'=e^2x^2-2x

Answer 10

F=c+{e^2x^2-2x dx

Answer 11

Therefore: y=e^-2x^2•(c+{e^2x^2-2x dx) →(1)

Answer 12

We know that, e^w= ∞summation lower n=0 w^n/n!

Answer 13

W_=2x^2

Answer 14

e^2x^2=1+2x^2+4x^4/2! +8x^6/3! +... →(2)

Answer 15

W_= -2x

Answer 16

e^-2x=1-2x+4x^2/2! -8x^3/3! +—... →(3)

Answer 17

Then e^2x^2-2x

Answer 18

=e^2x^2•e^-2x

Answer 19

100"9=90 n_2     how to get this answer  explain each detail ?       I'm not at school today!

Answer 20

=1+s(x)

Answer 21

Where s(x) is an infinite polynomial of the form:

Answer 22

∞ summation n=0 Kx^n

Answer 23

That is, there is not a constant term.

Answer 24

Therefore the integral in (1) is

Answer 25

{(1+s) dx

Answer 26

=x+{s(x) dx

Answer 27

=x+r(x)

Answer 28

Now r(x) has not also any constant term,

Answer 29

Thus r(0)=0

Answer 30

Then (1)becomes y(x)=e^-2x^2•(c+x+r(x))

Answer 31

→y(0)=e°•(c+0+0)_=1

Answer 32

→ c=1

Answer 33

Therefore the solution to the initial condition problem is given by,

Answer 34

y(x)=e^-2x^2•[1+{(EXP with ^upper(2x^2-2x))dx]