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By Pandey, Rajesh.

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Replacing X and Y in the solution so obtained by x-h and y-k [from (3)] respectively We can get the required solution is terms of x and y, the original variables. A Special Case When a:b =a':b' In this case we cannot solve the equations given by (4) above and the differential equation is of the form dy dx = ax + by + C kax + kby + C (5) In this case the differential equation is solve by putting v = ax + by (6) Differentiating both sides of (6) with respect to x, we get dv = a + b dy or dy = ! (dV _ a) dx dx dx b dx :.

From (I), the required solution is y (x + log x) + x cos x = C, where C is an arbitrary constant Example 17. S. 1996) Solution. Here M = X4 - 2xy2 + y4, N = - 2x2 Y+ 4xy3 - sin Y 24 Differential Equations of First Order and First Dezree :. -aM = -4xy + 4y 3 and -aN = -4xy + 4y 3 ax By aM = aN, h ence t h · . IS exact e gIven equation :. - ax By Regarding y as constant, fMdx = f(x 4 - 2xy2 + y4) dx (1) Integrating N with respect to y, we get f(-2x 2 y + 4 xy3 - sin y) dy = _x 2y2 + xy4 + cos Y omitting from this the term -x2y2and xytwhich are already occuring in (1) we get cosy :.

S. 1993) Solution. Here M = Y (1 + :. aM ay = aM ay =:> - (1 1). + -; - sm y an aN = -, ax ~) cos y and N = x + log x - x sin Y daNax =1 1 . y + -; - sm h ence t h · " IS exact e gIven equation Regarding y as constant, fMdx = f{y R1 (1 + ;) + cos y} dx ~) dx + cos Y fdx = y = Y(x + log x) + (cos y) x + (1) Also no new term is obtained by integrating N with respect to y. :. From (I), the required solution is y (x + log x) + x cos x = C, where C is an arbitrary constant Example 17. S. 1996) Solution.

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