Find the general solution of the given system dx/dt = -4x + 2y, dy/dt = -5/2 x + 2y

Solution:

dx/dt = -4x + 2y --- (1)

dy/dt = (-5/2)x + 2y --- (2)

Dividing (2) by (1) we get

dy/dx = [ (-5/2)x + 2y]/(-4x + 2y) --- (3)

To find the general solution let,

y = vx --- (4)

Differentiating the above w.r.t. X we have

dy/dx = v + xdv/dx --- (5)

Substituting (3) and (4) in equation (5) above we get,

[(-5/2)x + 2(vx)]/[(-4x + 2(vx)] = v + x(dv/dx)

⇒ (-5x + 4vx)/(-8x + 4vx) = v + xdv/dx

⇒ [(-5 + 4v)/(-8 + 4v)] - v = x(dv/dx)

⇒ (-5 + 4v + 8v - 4v²)/(-8 + 4v) = x(dv/dx)

⇒ (-5 + 12v -4v²)/(-8 + 4v) = x(dv/dx)

⇒ (-8 + 4v)dv/(-5 + 12v -4v²) = dx/x

⇒ (4v - 8)dv/(4v² - 12v + 5) = dx/x

⇒ (1/2)(8v - 16)dv/(4v² - 12v + 5) = -dx/x

Taking Integrals on both sides

⇒1/2\(\int\)(8v -12 - 4 )dv/(4v² - 12v + 5) = -\(\int\)dx/x

⇒ 1/2\(\int\)(8v - 12)dv/(4v² - 12v + 5) - 1/2\(\int\)4/(4v² - 12v + 5) = -LnIxI + c

⇒ (1/2 ) ln I 4v² - 12v + 5 I - (1/2)\(\int\)4dv/(4v² - 12v + 5) = - lnI x I + c

4v² - 12v + 5 is not convertible into a² + b² and hence the second term on the LHS does not have an integral solution

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Find the general solution of the given system dx/dt = -4x + 2y, dy/dt = -5/2 x + 2y

Summary:

There is no general solution of the given system dx/dt = -4x + 2y, dy/dt = -5/2 x + 2y