is homogeneous because both M( x,y) = x 2 – y 2 and N( x,y) = xy are homogeneous functions of the same degree (namely, 2). Recall that this will happen if  \(\begin{align}\frac{a}{b} = \frac{d}{e}.\end{align}\) How do we reduce the DE to a homogeneous one in such a case ? The first question that comes to our mind is what is a homogeneous equation? Solution: We substitute \(x \to X + h\) and \(y \to Y + k\) where h, k need to be determined : \[\frac{{dy}}{{dx}} = \frac{{dY}}{{dX}} = \frac{{(2Y - X) + (2k - h - 4)}}{{(Y - 3X) + (k - 3h + 3)}}\], \[\begin{array}{l}2k - h - 4 = 0\\k - 3h + 3 = 0\end{array}\]. = ∫( )Q I.F.× dx + C, where I.F. We now integrate this DE which is VS; the left-hand side can be integrated by the techniques described in the unit on Indefinite Integration. Finally, we substitute \(v = \frac{Y}{X}\) and, \[\begin{array}{l}X = x - 2\\Y = y - 3\end{array}\], \[\frac{{dy}}{{dx}} = f\left( {\frac{{ax + by + c}}{{dx + ey + f}}} \right)\], \[\begin{array}{l}ah + bk + c = 0\\dh + ek + f = 0\end{array}\]. Let \(\begin{align}\frac{a}{d} = \frac{b}{e} = \lambda \end{align}\) (say). A differential equation can be homogeneous in either of two respects.. A first order differential equation is said to be homogeneous if it may be written (,) = (,),where f and g are homogeneous functions of the same degree of x and y. Solve the DE \(\begin{align}\frac{{dy}}{{dx}} = \frac{{2y - x - 4}}{{y - 3x + 3}}.\end{align}\). The general solution of the homogeneous differential equation depends on the roots of the characteristic quadratic equation. Using the substitution \(Y = vX,\) and simplifying, we have (verify), \[\frac{{v - 3}}{{{v^2} - 5v + 1}}dv = \frac{{ - dX}}{X}\]. Section 7-2 : Homogeneous Differential Equations. In the preceding section, we learned how to solve homogeneous equations with constant coefficients. FREE Cuemath material for JEE,CBSE, ICSE for excellent results! Therefore, for nonhomogeneous equations of the form \(ay″+by′+cy=r(x)\), we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. Example 6: The differential equation . Thus, \[\frac{{ax + by + c}}{{dx + ey + f}} = \frac{{\lambda (dx + ey) + c}}{{dx + ey + f}}\], This suggests the substitution \(dx + ey = v,\) which’ll give, \[\begin{align}&\qquad \;\;d + e\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\\&\Rightarrow \quad \frac{{dy}}{{dx}} = \frac{1}{e}\left( {\frac{{dv}}{{dx}} - d} \right)\end{align}\], \[\begin{align} & \qquad \;\;\frac{1}{e}\left( {\frac{{dv}}{{dx}} - d} \right) = \frac{{\lambda v + c}}{{v + f}}\\&\Rightarrow \quad \frac{{dv}}{{dx}} = \frac{{\lambda ev + ec}}{{v + f}} + d\\ &\qquad \qquad= \frac{{(\lambda e + d)v + (ec + df)}}{{v + f}}\\&\Rightarrow \quad \frac{{(v + f)}}{{(\lambda e + d)v + ec + df}}dv = dx\end{align}\]. which is in VS form and hence can be solved. Therefore, for nonhomogeneous equations of the form \(ay″+by′+cy=r(x)\), we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. Instead, we will focus on special cases. Examples On Differential Equations Reducible To Homogeneous Form in Differential Equations with concepts, examples and solutions. (xiv) Another form of first order linear differential equation is dx dy + P 1 x = Q 1, where P 1 and Q 1 are constants or functions of y only. In this case, the change of variable y = ux leads to an equation of the form = (), which is easy to solve by integration of the two members. (Integrating Factor) = e∫Pdx. The next step is to investigate second order differential equations. There are the following options: Discriminant of the characteristic quadratic equation \(D \gt 0.\) Then the roots of the characteristic equations \({k_1}\) and \({k_2}\) are real and distinct. A simple, but important and useful, type of separable equation is the first order homogeneous linear equation: Definition 17.2.1 A first order homogeneous linear differential equation is one of the form $\ds \dot y + p(t)y=0$ or equivalently $\ds \dot y = -p(t)y$. We can note that f(αx,αy,αz) = (αx)2+(αy)2+(αz)2+… A function of form F(x,y) which can be written in the form k n F(x,y) is said to be a homogeneous function of degree n, for k≠0. For each equation we can write the related homogeneous or complementary equation: y′′+py′+qy=0. The general second order differential equation has the form \[ y'' = f(t,y,y') \label{1}\] The general solution to such an equation is very difficult to identify. A differential equation can be homogeneous in either of two respects. Thus, we use the substitution, \[\begin{align}&\qquad \;x + 2y = v\\& \Rightarrow \quad 1 + 2\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}\end{align}\], \[\begin{align} &\qquad \frac{1}{2}\left( {\frac{{dv}}{{dx}} - 1} \right) = \frac{{v - 1}}{{v + 1}}\\&\Rightarrow \quad \frac{{dv}}{{dx}} = \frac{{2v - 2}}{{v + 1}} + 1\\& \qquad \qquad = \frac{{3v - 1}}{{v + 1}}\\&\Rightarrow \quad \frac{{v + 1}}{{3v - 1}}dv = dx\\&\Rightarrow \quad \frac{1}{3}\left( {1 + \frac{4}{{3v - 1}}} \right)dv = dx\end{align}\], \[\frac{1}{3}\left( {v + \frac{4}{3}\ln (3v - 1)} \right) = x + {C_1}\], \[\begin{align}& \qquad \;\; x + 2y + \frac{4}{3}\ln (3x + 6y - 1) = 3x + {C_2}\\&\Rightarrow \quad y - x + \frac{2}{3}\ln (3x + 6y - 1) = C\end{align}\], Examples on Reducible to Homogeneous Form, Download SOLVED Practice Questions of Examples On Differential Equations Reducible To Homogeneous Form for FREE, Learn from the best math teachers and top your exams, Live one on one classroom and doubt clearing, Practice worksheets in and after class for conceptual clarity, Personalized curriculum to keep up with school.

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