IGNOU MTE 8 2026 - Find the integrating factor of the differential equation

Question

Find the integrating factor of the differential equation

$$(6xy - 3y^2 + 2y) \, dx + 2(x - y) \, dy = 0$

and hence solve it.

b) Solve the equation $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = x^m$ , for all positive integer values of m.
c) Solve the following IVP

$$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = -6 \sin 2x - 18 \cos 2x$

$$y(0) = 2, \quad y'(0) = 2.$

09 Jan 2026

Answer :

Word Count : 773

For the differential equation: [ (6xy - 3y^2 + 2y),dx + 2(x - y),dy = 0 ] let (M = 6xy - 3y^2 + 2y) and (N = 2(x - y)). Check if it is exact: [ \frac{\partial M}{\partial y} = 6x - 6y + 2, \quad \frac{\partial N}{\partial x} = 2 ] Since (\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}), it is not exact. Consider an integrating factor of the form (\mu = \mu(y)). Check (\frac{\partial}{\partial y}(\mu M) = \frac{\partial}{\partial x}(\mu N)): [ \mu' M + \mu \frac{\partial M}{\partial y} = \mu \frac{\partial N}{\partial x} \implies \frac{\mu'}{\mu} = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{2 - (6x - 6y + 2)}{6xy - 3y^2 + 2y} = \frac{-6x + 6y}{6xy - 3y^2 + 2y} = \frac{6(y-x)}{6xy - 3y^2 + 2y} ] Factor 3 in denominator: [ \frac{\mu'}{\mu} = \frac{2(y-x)}{2y(3x - y + 1)} \text{ is not solely a function of } y ] Try (\mu = \mu(x)). Then: [ \frac{\mu'}{\mu} = \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{(6x - 6y + 2) - 2}{2(x-y)} = \frac{6x - 6y}{2(x-y)} = 3 ] Hence (\mu(x) = e^{\int 3 dx} = e^{3x}). Multiply entire equation by (e^{3x}): [ e^{3x}(6xy - 3y^2 + 2y),dx + e^{3x} 2(x-y),dy = 0 ] Now exact: let (F(x,y)) such that (F_x = e^{3x}(6xy - 3y^2 + 2y), \ F_y = e^{3x} 2(x-y)). Integrate (F_x) w.r.t (x): [ F = \int e^{3x}(6xy - 3y^2 + 2y) dx = 2y e^{3x} x - y^2 e^{3x} + \frac{2}{3} y e^{3x} + \phi(y) = e^{3x}(3xy - 3y^2 + \ldots) ] Check derivative w.r.t (y), adjust (\phi(y)). Finally solution: [ e^{3x}(3xy - 3y^2 + y) = C ] --- For (x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = x^m): divide by (x^2) [ \frac{d^2y}{dx^2} + \frac{1}{x}\frac{dy}{dx} + \frac{1}{x^2}y = x^{m-2}

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