What is the general solution of the differential equation # x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x #?

Question

Emilia Aguilar · Accepted Answer

# y = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + (5sinlnx)/(61x) + (6coslnx)/(61x) #

We have: # x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x # ..... [A] This is a Euler-Cauchy Equation (the power of #x# is the same as the degree of the differential in every occurrence of their product) which is typically solved via a change of variable. Consider the substitution: # x = e^t => xe^(-t)=1# Next up, we have #dy/dx = e^(-t)dy/dt#, and, #(d^2y)/(dx^2)=((d^2y)/(dt^2)-dy/dt)e^(-2t)# When we replace the original DE [A], we obtain: # (e^t)^2 ((d^2y)/(dt^2)-dy/dt)e^(-2t) - 3(e^t) e^(-t)dy/dt+y=sin(loge^t)/((e^t)) # # :. (d^2y)/(dt^2)-dy/dt - 3dy/dt+y = e^(-t)sint # # :. (d^2y)/(dt^2) - 4dy/dt+y = e^(-t)sint # ..... [B] This is now a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation. Complementary Role The homogeneous equation that goes with it is: # (d^2y)/(dt^2) - 4dy/dt+y = 0 # ..... [C] whose Auxiliary Equation is: # m^2-4m+1 = 0# This quadratic equation can be solved, and the two distinct real solutions that result are: # m=2+-sqrt(3) # Consequently, the solution to the homogeneous equation [C] is: # y_c = Ae^((2-sqrt(3))t) + Be^((2+sqrt(3))t)# Specific Resolution A likely solution for this particular equation [B] is as follows: # y = ae^(-t)sint + be^(-t)cost # # \ \ = e^(-t)(asint + bcost) # Where #a# and #b# are constants to be determined. Let us assume the above solution works, in which case be differentiating wrt #x# we have: # y' \ \= e^(-t)(acost - bsint) - e^(-t)(asint + bcost) # # \ \ \ \ \= e^(-t)( (a-b)cost - (a+b)sint ) # # y'' = e^(-t)(-(a-b)sint - (a+b)cost ) - e^(-t)((a-b)cost - (a+b)sint ) # # \ \ \ \ \ = e^(-t)(2bsint -2acost) # Substituting into the Differential Equation #[B]# we get: # e^(-t)(2bsint -2acost) - 4e^(-t)( (a-b)cost - (a+b)sint ) + e^(-t)(asint + bcost) = e^(-t)sint # Equating coefficients of #cos(x)# and #sin(x)# we get: #cos(x): -2a - 4(a-b) + b = 0 # #sin(x): 2b +4 (a+b) + a = 1 # When we solve concurrently, we obtain: # a=5/61 # and #b=6/61# Thus, we arrive at the specific solution: # y_p = 5/61e^(-t)sint + 6/61e^(-t)cost # Overall Resolution which ultimately results in the GS of [B^ # y(t) = y_c + y_p # # \ \ \ \ \ \ \ = Ae^((2-sqrt(3))t) + Be^((2+sqrt(3))t) + 5/61e^(-t)sint + 6/61e^(-t)cost# Now, in the beginning, we employed a variable change: # x = e^t => t=lnx # Restoring this variable change gives us the following: # y = Ae^((2-sqrt(3))lnx) + Be^((2+sqrt(3))lnx) + 5/61e^(-lnx)sinlnx + 6/61e^(-lnx)coslnx# # :. y = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + 5/61e^(ln(1/x))sinlnx + 6/61e^(ln(1/x))coslnx# # \ \ \ \ \ \ = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + 5/61 (1/x)sinlnx + 6/61 (1/x)coslnx # # \ \ \ \ \ \ = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + (5sinlnx)/(61x) + (6coslnx)/(61x) # This is [A]'s General Solution.

Isabella Ackerman · Answer

undefined To find the general solution of the given differential equation $x^2 \frac{{d^2y}}{{dx^2}} - 3x \frac{{dy}}{{dx}} + y = \frac{{\sin(\log x)}}{x}$, follow these steps:

1. First, we'll rewrite the equation in a standard form, which is a second-order linear homogeneous differential equation:
$$x^2 \frac{{d^2y}}{{dx^2}} - 3x \frac{{dy}}{{dx}} + y = 0$$
$$x^2 y'' - 3xy' + y = 0$$

2. Next, we'll find the general solution to the homogeneous equation $x^2 y'' - 3xy' + y = 0$.

3. To solve this, we'll assume the solution has the form $y = x^m$, where $m$ is a constant.

4. Substitute $y = x^m$ into the homogeneous differential equation:
$$x^2 (m(m-1)x^{m-2}) - 3x(mx^{m-1}) + x^m = 0$$

5. Simplify the equation:
$$m(m-1)x^m - 3mx^m + x^m = 0$$
$$m(m - 1 - 3 + 1)x^m = 0$$
$$m(m - 3)x^m = 0$$

6. The characteristic equation is $m(m - 3) = 0$.

7. Solve the characteristic equation to find the roots:
$$m(m - 3) = 0$$
$$m = 0, 3$$

8. Since we have distinct real roots, the general solution to the homogeneous differential equation is:
$$y = C_1 x^0 + C_2 x^3$$
$$y = C_1 + C_2 x^3$$

9. Now, we'll find a particular solution to the non-homogeneous equation using the method of undetermined coefficients or variation of parameters.

10. Given the non-homogeneous term $\frac{{\sin(\log x)}}{x}$, we'll guess a particular solution of the form $y_p = A \sin(\log x) + B \cos(\log x)$.

11. Substitute $y_p = A \sin(\log x) + B \cos(\log x)$ into the non-homogeneous differential equation and solve for $A$ and $B$.

12. Once we have $A$ and $B$, the general solution to the non-homogeneous differential equation is:
$$y = C_1 + C_2 x^3 + A \sin(\log x) + B \cos(\log x)$$

13. Therefore, the general solution to the given differential equation is the sum of the general solution to the homogeneous equation and the particular solution to the non-homogeneous equation:
$$y = C_1 + C_2 x^3 + A \sin(\log x) + B \cos(\log x)$$