Answer:
Explanation:
The Bernoulli differential equation is given by:
y' + p(x)y = q(x)y^n
where n is a constant, not equal to 0 or 1. In this case, we have:
y' + (x-3)/x y = (y+x)/x^2 y^2
To solve this equation, we can use the substitution:
u = y^(1-n)
So that:
y = u^(1/(1-n))
y' = (1/(1-n))u^(1/(1-n)-1)u'
Substituting these expressions into the Bernoulli equation and simplifying, we get:
(1/(1-n))u^(1/(1-n)-1)u' + (x-3)/x u^(1/(1-n)) = (u^(1/(1-n))+x)/x^2 (u^(2/(1-n)))
Multiplying both sides by (1-n)u^(1/(1-n)) and simplifying, we get:
(1-n)u^(1/(1-n))u' + (x-3)/x (1-n)u = (u^(1/(1-n))+x)/x^2 (u^(2/(1-n)+1))
The left-hand side is the derivative of (1-n)u^(1/(1-n)) with respect to u, so we can write:
d/dx[(1-n)u^(1/(1-n))] = (u^(1/(1-n))+x)/x^2 (u^(2/(1-n)+1))
Integrating both sides with respect to u, we get:
(1-n)u^(1/(1-n)) = (1/(2/(1-n)+1))u^(2/(1-n)+1) + C
Substituting u = y^(1-n), we get:
(1-n)y = (1/(2/(1-n)+1))y^(2/(1-n)+1) + C
To find the particular solution to the non-homogeneous equation, we can use the method of undetermined coefficients and guess that the particular solution has the form YP = ax + b. Then:
YP' = a
YP^2 = a^2x^2 + 2abx + b^2
Substituting these expressions into the non-homogeneous equation and simplifying, we get:
2a^2x^2 + (2ab - a)x + (b^2 - b) = 0
For this equation to hold for all x, we must have:
2a^2 = 0
2ab - a = 1
b^2 - b = 0
The first equation gives a = 0, and substituting this into the second equation gives b = 1. Therefore, the particular solution is:
YP = 2x
To find the general solution to the non-homogeneous equation, we add the particular solution to the homogeneous solution we found earlier:
y = (1/(1-n))u^(1/(1-n)) + 2x
Substituting u = y^(1-n), we get:
y = (1/(1-n))(y^(1-n))^(1/(1-n)) + 2x
Simplifying:
y = ((1-n)y)^(-1/n) + 2x
Therefore, the general solution to the non-homogeneous equation is:
y = ((1-n)y)^(-1/n) + 2x
Note that this solution is only valid for n not equal to 1 or 0, since those cases correspond to the linear and separable differential equations, respectively.