To solve the differential equation xy” + 4y’ = x^4 using the method of variation of parameters, we first need to rewrite the equation in a standard form. The equation can be rearranged as:
y” + rac{4}{x}y’ = x^3
Next, we identify the homogeneous part of the equation:
y” + rac{4}{x}y’ = 0
We can solve this homogeneous equation by finding its characteristic equation or by using another method. The solution to the homogeneous equation typically takes the form:
y_h = C_1 y_1 + C_2 y_2
where C_1 and C_2 are constants, and y_1 and y_2 are the linearly independent solutions we find. In this case, we can use the method of undetermined coefficients or simply guess a solution based on the structure of the equation.
For this equation, we assume solutions of the form:
y_1 = x^n
By substituting different values for n, we derive:
- For n = 0:
y_1 = 1 - For n = 1:
y_2 = ext{ln}(x)
Thus, the complementary solution (homogeneous solution) can be expressed as:
y_h = C_1 + C_2 ext{ln}(x)
Next, we move on to the particular solution using the method of variation of parameters, which involves setting:
y_p = u_1 y_1 + u_2 y_2
where u_1 and u_2 are functions that we need to determine. The formulas for u_1 and u_2 in the method of variation of parameters are given by:
u_1 = - rac{y_2 g}{W}, u_2 = rac{y_1 g}{W}
Here, g = x^3 and W is the Wronskian of y_1 and y_2.
The Wronskian can be calculated as:
W = y_1 y_2′ – y_2 y_1′ = 1 imes rac{1}{x} – ext{ln}(x) imes 0 = rac{1}{x}
With these, we substitute to find:
u_1 = - rac{ ext{ln}(x) x^3}{ rac{1}{x}} = -x^4 ext{ln}(x)
u_2 = rac{1 imes x^3}{ rac{1}{x}} = x^4
Thus:
y_p = -x^4 ext{ln}(x) + x^4
To find the general solution, we combine the homogeneous and particular solutions:
y = y_h + y_p = C_1 + C_2 ext{ln}(x) – x^4 ext{ln}(x) + x^4
This is the general solution to the given differential equation. Always remember to check your solution by substitution back into the original equation to confirm it satisfies all conditions!