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.

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