Ch 9 (8 9 09)

1 Overview(i)An equation involving derivative (derivatives) of the dependent variable withrespect to independent variable (variables) is called a differential equation.(ii)A differential equation involving derivatives of the dependent variable withrespect to only one independent variable is called an ordinary differentialequation and a differential equation involving derivatives with respect to morethan one independent variables is called a partial differential equation.(iii)Order of a differential equation is the order of the highest order derivativeoccurring in the differential equation.(iv)Degree of a differential equation is defined if it is a polynomial equation in itsderivatives.(v)Degree (when defined) of a differential equation is the highest power (positiveinteger only) of the highest order derivative in it.

2 (vi)A relation between involved variables, which satisfy the given differentialequation is called its solution. The solution which contains as many arbitraryconstants as the order of the differential equation is called the general solutionand the solution free from arbitrary constants is called particular solution.(vii)To form a differential equation from a given function, we differentiate thefunction successively as many times as the number of arbitrary constants in thegiven function and then eliminate the arbitrary constants.(viii) The order of a differential equation representing a family of curves is same asthe number of arbitrary constants present in the equation corresponding to thefamily of curves.(ix) Variable separable method is used to solve such an equation in which variablescan be separated completely, , terms containing x should remain with dx andterms containing y should remain with EQUATIONS20/04/2018180 MATHEMATICS(x) A function F (x, y) is said to be a homogeneous function of degree n ifF ( x, y )= n F (x, y) for some non-zero constant.

3 (xi) A differential equation which can be expressed in the form dydx= F (x, y) ordxdy = G (x, y), where F (x, y) and G (x, y) are homogeneous functions of degreezero, is called a homogeneous differential equation.(xii) To solve a homogeneous differential equation of the type dydx = F (x, y), we makesubstitution y = vx and to solve a homogeneous differential equation of the typedxdy = G (x, y), we make substitution x = vy. (xiii) A differential equation of the form dydx + Py = Q, where P and Q are constants orfunctions of x only is known as a first order linear differential equation. Solutionof such a differential equation is given by y ( ) = () + C, (Integrating Factor) = Pdxe .(xiv) Another form of first order linear differential equation is dxdy + P1x = Q1, whereP1 and Q1 are constants or functions of y only.

4 Solution of such a differentialequation is given by x ( ) = ()1Q + C, where = 1 Pdye . Solved ExamplesShort Answer ( )Example 1 Find the differential equation of the family of curves y = Ae2x + y = Ae2x + 2x20/04/2018 differential EQUATIONS 181dydx = 2Ae2x 2 2x and 22dydx = 4Ae2x + 4Be 2xThus22dydx = 4y ,22dydx 4y = 2 Find the general solution of the differential equation dydx= yx dyy = dxx dyy = dxx logy = logx + logc y = cxExample 3 Given that dydx= yex and x = 0, y = e. Find the value of y when x = dydx= yex dyy = xedx logy = ex + cSubstituting x = 0 and y = e,we get loge = e0 + c, , c = 0 ( loge = 1)Therefore, log y = , substituting x = 1 in the above, we get log y = e y = 4 Solve the differential equation dydx + yx= The equation is of the type +P= Qdyydx, which is a linear = 1dxx = elogx = , solution of the given differential equation is20/04/2018182 = 2xxdx , yx = 44xc+Hencey = 34xcx+.

5 Example 5 Find the differential equation of the family of lines through the Let y = mx be the family of lines through origin. Therefore, dydx= mEliminating m, we get y = dydx. x or xdydx y = 6 Find the differential equation of all non-horizontal lines in a The general equation of all non-horizontal lines in a plane isax + by = c, where a , dxabdy+= , differentiating both sides y, we get22dxady = 0 22dxdy= 7 Find the equation of a curve whose tangent at any point on it, differentfrom origin, has slope yyx+.Solution Given dyyydxx=+= 11yx + 11dydxyx =+ Integrating both sides, we getlogy = x + logx + c logyx = x + c20/04/2018 differential EQUATIONS 183 yx= ex + c = yx= k.

6 Ex y = kx . Answer ( )Example 8 Find the equation of a curve passing through the point (1, 1) if theperpendicular distance of the origin from the normal at any point P(x, y) of the curveis equal to the distance of P from the x Let the equation of normal at P(x, y) be Y y = () X dxxdy, ,Y + Xdxdy dxyxdy + = (1)Therefore, the length of perpendicular from origin to (1) is21dxyxdydxdy+ + ..(2)Also distance between P and x-axis is |y|. Thus, we get21dxyxdydxdy+ + = |y| 2dxyxdy + = 221dxydy + ()22 20dxdxxyxydydy += 0dxdy=ordxdy= 222 xyyx20/04/2018184 MATHEMATICSCase I: dxdy = 0 dx = 0 Integrating both sides, we get x = k, Substituting x = 1, we get k = , x = 1 is the equation of curve (not possible, so rejected).

7 Case II:dxdy= 222222xydyyxdxxyyx . Substituting y = vx, we get22222dvvxxvxdxvx += = = 2(1)2vv + 221vdxdvxv =+Integrating both sides, we getlog (1 + v2) = logx + logc log (1 + v2) (x) = log c (1 + v2) x = c x2 + y2 = cx. Substituting x = 1,y = 1,we getc = ,x2 + y2 2x = 0is the required 9 Find the equation of a curve passing through 1,4 if the slope of thetangent to the curve at any point P (x, y) is 2cosyyxx .Solution According to the given condition2cosdyyydxxx= .. (i)This is a homogeneous differential equation. Substituting y = vx, we getv + x dvdx = v cos2v dvxdx = cos2v20/04/2018 differential EQUATIONS 185 sec2v dv = dxx tan v = logx + c tan logyxcx+=.

8 (ii) Substituting x = 1, y = 4 , we get. c = 1. Thus, we gettan yx + log x = 1, which is the required 10 Solve 2dyxxydx = 1 + cos yx , x 0 and x = 1, y = 2 Solution Given equation can be written as2dyxxydx = 2cos2 2yx , x 0. 2212cos2dyxxydxyx = 22sec212ydyxxxydx = Dividing both sides by x3 , we get223sec122ydyxyxdxxx = 31tan2dydxxx = Integrating both sides, we get21tan22ykxx =+ .20/04/2018186 MATHEMATICSS ubstituting x = 1, y = 2 , we getk = 32, therefore, 213tan222yxx = + is the required 11 State the type of the differential equation for the ydx = 22xy+dx and solve Given equation can be written as xdy = ()22xyydx++, ,22xyydydxx++=.

9 (1)Clearly RHS of (1) is a homogeneous function of degree zero. Therefore, the givenequation is a homogeneous differential equation. Substituting y = vx, we get from (1)222xvxvxdvvxdxx+++= 21dvvxvvdx+=++21dvxvdx=+ 21dvdxxv=+.. (2)Integrating both sides of (2), we getlog (v + 21v+) = logx + logc v + 21v+ = cx yx + 221yx+ = cx y + 22xy+ = cx220/04/2018 differential EQUATIONS 187 Objective Type QuestionsChoose the correct answer from the given four options in each of the Examples 12 to 12 The degree of the differential equation 23221dydydxdx += is(A) 1(B) 2(C) 3(D) 4 Solution The correct answer is (B).Example 13 The degree of the differential equation2222223logdydydyxdxdxdx += is(A) 1(B) 2(C) 3(D) not definedSolution Correct answer is (D).

10 The given differential equation is not a polynomialequation in terms of its derivatives, so its degree is not 14 The order and degree of the differential equation22221dydydxdx += respectively, are(A) 1, 2(B) 2, 2(C) 2, 1 (D) 4, 2 Solution Correct answer is (C).Example 15 The order of the differential equation of all circles of given radius a is:(A) 1(B) 2(C) 3(D) 4 Solution Correct answer is (B). Let the equation of given family be(x h)2 + (y k)2 = a2 . It has two orbitrary constants h and k. Threrefore, the order ofthe given differential equation will be 16 The solution of the differential equation 2. dyxydx= 3 represents a family of(A) straight lines (B) circles(C) parabolas(D) ellipses20/04/2018188 MATHEMATICSS olution Correct answer is (C).