By Peter V. O'Neil
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Additional resources for Advanced Engineering Mathematics, International Student Edition
16). 16). This solution was “lost”, or at least not found, in using the integrating factor as a method of solution. However, y = 3 is not a singular solution because we can include it in the expression y = 3 + Kx by allowing K = 0. 16) is y = 3 + Kx, with K any real number. 1 37 Separable Equations and Integrating Factors We will point out a connection between separable equations and integrating factors. The separable equation y = A x B y is in general not exact. To see this, write it as A x B y −y = 0 so in the present context we have M x y = A x B y and N x y = −1.
10. x2 y = x2 + y2 1 2 11. y = − y2 + y x x 12. x3 y = x2 y − y3 13. y = −e−x y2 + y + ex 3 2 14. y + y = y2 x x 15. Consider the differential equation y =F 45 ax + by + c dx + ey + r in which a, b, c, d, e, and r are constants and F is a differentiable function of one variable. (a) Show that this equation is homogeneous if and only if c = r = 0. (b) If c and/or r is not zero, this equation is called nearly homogeneous. Assuming that ae − bd = 0, show that it is possible to choose constants h and k so that the transformation X = x + h Y = y + k converts this nearly homogeneous equation into a homogeneous one.
Thus, as we have seen before, when we perform manipulations on a differential equation, we must be careful that solutions have not been overlooked. 17) may have other solutions as well. Now to the point. A homogeneous equation is always transformed into a separable one by the transformation y = ux. To see this, compute y = u x + x u = u x + u and write u = y/x. Then y = f y/x becomes u x+u = f u We can write this as 1 du 1 = f u − u dx x or, in differential form, 1 1 du = dx f u −u x and the variables (now x and u) have been separated.
Advanced Engineering Mathematics, International Student Edition by Peter V. O'Neil