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.
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34 Answers

Let Y,h: Y'h+4xY,h=0
by Level 12 User (101k points)
→{dyh/Yh={-4xdx
by Level 12 User (101k points)
→luY,h=-2x^2+c ~ upper
by Level 12 User (101k points)
Y,h= c•e^-2x^2
by Level 12 User (101k points)
Now if F_=F(x):
by Level 12 User (101k points)
Y=F•e^-2x^2
by Level 12 User (101k points)
Y'=(F'-4xF)e-2x^2
by Level 12 User (101k points)
And by substitution to the given
by Level 12 User (101k points)
F'=e^2x^2-2x
by Level 12 User (101k points)
F=c+{e^2x^2-2x dx
by Level 12 User (101k points)
Therefore : Y=e^-2x^2•(c+{e^2x^2-2x dx) →(1)
by Level 12 User (101k points)
We know that, e^w=∞ upper summation n=0 w^n/n!
by Level 12 User (101k points)
W_=2x^2
by Level 12 User (101k points)
e^2x^2=1+2x^2+4x^4/2! +8x^6/3! +... →(2)
by Level 12 User (101k points)
W_= -2x
by Level 12 User (101k points)
e^-2x=1-2x+ 4x^2/2!-8x^3/3! +—... →(3)
by Level 12 User (101k points)
Then e^2x^2-2x
by Level 12 User (101k points)
=e^2x^2•e^-2x
by Level 12 User (101k points)
=1+s(x)
by Level 12 User (101k points)
Where s(x) is an infinite polynomial of the form
by Level 12 User (101k points)
∞summation n=1 kx^n
by Level 12 User (101k points)
That is, there is not a constant term
by Level 12 User (101k points)
Therefore the integral in (1) is
by Level 12 User (101k points)
{(1+s) dx= x+ {s(x) dx= x+r(x)
by Level 12 User (101k points)
Now r(x) has not also any constant term
by Level 12 User (101k points)
This r(0)=0
by Level 12 User (101k points)
Then (1) becomes,
by Level 12 User (101k points)
Y(x)= e^-2x^2•(c+x+r(x))
by Level 12 User (101k points)
Y(0)=e°•(c+0+0)_=1
by Level 12 User (101k points)
→ c= 1
by Level 12 User (101k points)
Therefore:
by Level 12 User (101k points)
The solution to the initial condition problem
by Level 12 User (101k points)
Is given by,
by Level 12 User (101k points)
y(x)=e^-2x^2•[1+{(^EXP(2x^2-2x))dx]
by Level 12 User (101k points)

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