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Find the general solution of the given differential equation.

(x + 1) dy/dx + (x + 2)y = 2xe?x

y=

Give the largest interval I over which the general solution is defined. (Think about the implications of any singular points.)

(??, ?)

(0, 1)

(?1, ?)

(0, ?)

(??, 1)

Sagot :

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).

Learn more about integration -

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