(x^2)y"+3xy'=(x^2)

Anonymous

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(x^2)y"+3xy'=(x^2)

 

Answer No.1

GamaMaster0

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We divide by x^2 y''+3/xy'=1 this is a linear equation the integrating factor is (I=e^{int 3/x dx}=e^{3ln(x)}=x^3) and we multiply the equation by I (x^3y''+3x^2y'=x^3) ((x^3y')'=x^3) (x^3y'=int x^3dx=x^4/4+C) (y'=x/4+C/x^3) (y=x^2/8+D/x^2+E)

Answer No.2

Anonymous

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We get the homogeneous solution to differential equations of this sort by assuming the solution is x^n. This is what we use when the power of x equals the order of the derivative associated with it. Then, y' = nxn-1 and y'' = n(n-1)xn-2 Then, substituting into the above, we have x2(n(n-1)x^n-2) + 3x(nx^n-1) = 0 Factoring, we have (n(n-1) +3n)x^n = 0 As x^n is not equal to 0, (n2 +2n) = 0 n(n+2)= 0 n = 0, -2 x^0 = 1 Thus, our homogeneous solution is c1 + c2/x2 Next, we get the specific solution. We assume the solution is c3x2 (this should work, because as we have already shown, we will get this degree of x back from each term. Then, (x2)y"+3xy'= 2*1 * c3x2 + 2* c3x2 = 4c3x2 4c3 = 1 c3 = 1/4 Thus, the general solution is c1 + c2/x2 + 1/4x2

Answer No.3

Anonymous

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When factoring, you need to reverse the FOIL process (First, Inside, Outside, Last). Two factors will multiply to the starting expression. Start with the first value (2x^2). Two values that will multiply to this are 2x and x. So.... (2x + )(x + ) Two values that will multiply to y^2 are y and y. So.... (2x + y)(x + y) Multiply this out using the FOIL method... 2x^2 + yx + 2xy + y^2 2x^2 + 3xy + y^2 Note that the inside value is +3xy not -3xy. You need to play with signs. Try (2x - y)(x - y) 2x^2 - yx - 2xy + y^2 2x^2 - 3xy + y^2 That is your starting expression, so the answer is... (2x - y)(x - y)

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