Solve the initial value problem dy/dx + 4xy = e−2x with y(0) = 1

Question

Solve the initial value problem dy/dx + 36y = 0 with y(0) = 0 and dy/dx|x=0 = 5

If sin x, cos x, x sin x and x cos x are four possible solutions to a fourth order linear differential equation

calculate the Wronskian and check whether these four solutions are independent

Solve the boundary value problem d2y/dx2 + 2 dy/dx + y = 0 with y(0) = 1, y(log 3) = 2

Solve the initial value problem d3y/dx3 + 6 d2y/dx2 + 9 dy/dx = 0 with y(0) = 0, dy/dx|x=0 = 2, d2y/dx2 |x=0 = 1.

Obtain the exact differential equation for which the solution has the form x2 + y2 = 1. Show that theobtained differential equation is exact.

Answer 1

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

Answer 2

→{dyh/Yh={-4xdx

Answer 3

→luY,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=∞ upper summation 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

=1+s(x)

Answer 20

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

Answer 21

∞summation n=1 kx^n

Answer 22

That is, there is not a constant term

Answer 23

Therefore the integral in (1) is

Answer 24

{(1+s) dx= x+ {s(x) dx= x+r(x)

Answer 25

Now r(x) has not also any constant term

Answer 26

This r(0)=0

Answer 27

Then (1) becomes,

Answer 28

Y(x)= e^-2x^2•(c+x+r(x))

Answer 29

Y(0)=e°•(c+0+0)_=1

Answer 30

→ c= 1

Answer 31

Therefore:

Answer 32

The solution to the initial condition problem

Answer 33

Is given by,

Answer 34

y(x)=e^-2x^2•[1+{(^EXP(2x^2-2x))dx]