Solve
x^3y^'''+5x^2y^''+7xy^'+8y=t+e^(-t)
well, the solution of the homogeneous equations is
y = c1 cos(2lnx) + c2 sin(2lnx) + c3/x^2
Not sure what to do with the t's, though.
steve how u got that solution?
To solve the given differential equation, we can use the method of undetermined coefficients. This method assumes that the particular solution has a similar form to the non-homogeneous term. We'll start by finding the general solution to the homogeneous equation and then find a particular solution for the non-homogeneous term.
The homogeneous equation is obtained by setting the right-hand side (RHS) of the equation to zero:
x^3y'''+5x^2y''+7xy'+8y = 0
To solve this homogeneous equation, we can assume a solution of the form y(x) = x^r.
Plugging this solution into the equation gives us:
r(r-1)(r-2)x^r + 5r(r-1)x^r + 7rx^r + 8x^r = 0
Combining like terms:
(r^3 - 3r^2 + 5r + 8) x^r = 0
For this equation to hold for all x ≠ 0, the expression in parentheses must be zero:
r^3 - 3r^2 + 5r + 8 = 0
We can solve this cubic equation by factoring or using the Rational Root Theorem. In this case, there are no simple rational roots. Therefore, we'll use numerical methods or a calculator to find the approximate solutions:
r = -1.511, 1.106 ± 1.576i
The general solution to the homogeneous equation is then:
y_h(x) = c1*x^(-1.511) + c2*x^(1.106)*cos(1.576 ln(x)) + c3*x^(1.106)*sin(1.576 ln(x))
Now, let's find a particular solution for the non-homogeneous equation:
We can assume a particular solution of the form:
y_p(x) = Ax + B + C*e^(-t)
Differentiating y_p(x) with respect to x three times and substituting it back into the original equation, we get:
x^3(-C*e^(-t)) + 5x^2(0) + 7x(A - C*e^(-t)) + 8(Ax + B + C*e^(-t)) = t + e^(-t)
Simplifying the equation, we have:
- C*x^3*e^(-t) + 7Ax + 8Ax + 8B + (7C - e^(-t))x = t + e^(-t)
Comparing coefficients on both sides:
- C = 0
7A + (7C - 1) = 1
8A + 8B = 0
From the first equation, we find that C must be zero.
From the second equation, substituting C = 0, we find A = 1/7.
From the third equation, substituting A = 1/7, we find B = -1/7.
So, the particular solution is:
y_p(x) = (1/7)x - 1/7
Therefore, the general solution to the non-homogeneous differential equation is:
y(x) = y_h(x) + y_p(x)
= c1*x^(-1.511) + c2*x^(1.106)*cos(1.576 ln(x)) + c3*x^(1.106)*sin(1.576 ln(x)) + (1/7)x - 1/7
where c1, c2, and c3 are arbitrary constants determined by any initial or boundary conditions given in the problem.