Answer :
The general solution of the given differential equation is y = (2x[tex] {e}^{(xy)} [/tex] × (x+1) + [tex] {2x}^{ {2e}^{(xy)} } [/tex] + C) / (x+1)(x+2) and the largest interval is (-infinity, -2) U (-1, infinity).
The general solution of the mentioned differential equation will be calculated as follows -
Firstly rewriting the differential equation in standard form:
(x+1) dy/dx + (x+2)y = 2x[tex] {e}^{(xy)} [/tex]
Calculating the integrating factor to make the LHS as a total derivative:
As known, formula for integrating factor is:
mu = [tex] {e}^{(integral of (x+2)/(x+1) dx)} [/tex]
= [tex] {e}^{(ln(x+1) + ln(x+2))} [/tex]
= (x+1)(x+2)
Now multiplying both sides of the equation by the integrating factor:
(x+1)(x+2) × (x+1) dy/dx + (x+1)(x+2) × (x+2)y = 2x[tex] {e}^{(xy)} [/tex] × (x+1)(x+2)
LHS will be a total derivative. Rearrange the equation for the same -
d/dx((x+1)(x+2)y) = 2x[tex] {e}^{(xy)} [/tex] × (x+1)(x+2)
Integrating both sides of the equation:
(x+1)(x+2)y = integral of 2x[tex] {e}^{(xy)} [/tex] × (x+1)(x+2) dx
= 2x[tex] {e}^{(xy)} [/tex] × (x+1) + [tex] {2x}^{ {2e}^{(xy)} } [/tex] + C
Now find the equation for y:
y = (2x[tex] {e}^{(xy)} [/tex] × (x+1) + [tex] {2x}^{ {2e}^{(xy)} } [/tex] + C) / (x+1)(x+2)
Equation for y is the general solution of the differential equation where constant C is the arbitrary constant of integration.
The general solution is not defined at the position x=-1 and x=-2. Hence, the largest interval over which the general solution is defined is (-infinity, -2) U (-1, infinity).
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