Differential Equation Notes

Differential Equation It is not certain that everything is uncertain ...........Pascal, Blaise Introduction : An equation involving independent and dependent variables and the derivatives of the dependent variables is called a differential equation. There are two kinds of differential equation: 1.1Ordinary Differential Equation : If the dependent variables depend on one independent variable x, then the differential equation is said to be ordinary. dy dz + = y + z, for example dx dx d3 y dy dy + xy = sin x , + y = ex , +2 3 dx dx dx k 1.2 d2 y dx 2 3/2   dy 2  = 1 +      dx   , y=x dy +k dx   dy 2  1 +      dx   Partial differential equation : If the dependent variables depend on two or more indepen dent variables, then it is known as partial differential equation ∂ z ∂2z ∂2z ∂2z = ax, for example y2 +y + = 0 ∂ x ∂ y2 ∂ x2 ∂ y2 Order and Degree of a Differential Equation: 2.1 Order : Order is the highest differential appearing in a differential equation. 2.2 Degree : It is determined by the highest degree of the highest order derivative present in it after the differential equation is cleared of radicals and fractions so far as the derivatives are concerned. Note : In the differential equation, all the derivatives should be expressed in the polynomial form n n n  dm y  1  dm−1y  2  dy  k f1 (x, y)  m  + f2 (x, y)  m−1  + ........ fk(x, y)   = 0  dx   dx   dx  The above differential equation has the order m and degree n1. Example # 1:Find the order & degree of following differential equations. (i) (iii) 1/ 4 6   dy    = + y  dx   dx 2      dy d2 y  tan–1  x =y +  dx dx 2    d2 y (ii)  d3 y  dy 2  y = loge  3 +      dx    dx (iv) ey′″ – xy″ + y = 0 ∴ order = 2, degree = 4 ∴ order = 3, degree = 1 ∴ order = 2, degree = 1 4 Solution : (i) 6  d2 y   dy   2  = y +    dx   dx  2 (ii) (iii) d3 y  dy  +   = ey 3 dx  dx  d2 y dx 2 +x d3 y (iv) 3 e dx –x dy = tan y dx d2 y +y=0 dx 2 equation can not be expressed as a polynomial in differential coefficients, so degree is not applicable but order is 3. Differential Equation Self Practice Problems : (1) (i) Find order and degree of the following differential equations. dy dy dy 1 +y= (ii) – 7x = 0 –4 dy dx dx dx dx 2 (iii) Ans.  dy 1/ 2  d2 y   = + y  dx 2  dx   (1) (i) order = 1, degree = 2 (iii) order = 2, degree = 2 (ii) order = 1, degree = 2 Formation of Differential Equation: Differential equation corresponding to a family of curve will have : (a) (b) Order exactly same as number of essential arbitrary constants in the equation of curve. No arbitrary constant present in it. The differential equation corresponding to a family of curve can be obtained by using the following steps: (i) Identify the number of essential arbitrary constants in equation of curve. NOTE : If arbitrary constants appear in addition, subtraction, multiplication or division, then we can club them to reduce into one new arbitrary constant. (ii) Differentiate the equation of curve till the required order. (iii) Eliminate the arbitrary constant from the equation of curve and additional equations obtained in step (ii) above. Example # 2 : Form a differential equation of family of straight lines passing through (0,2) Solution : Family of straight lines passing through (0,2) is y = mx + 2 where’m’ is a parameter. Differentiating w.r.t. x dy =m dx Eliminating ‘m’ from both equations, we obtain dy +2 which is the required differential equation. y=x dx Example # 3 :Form a differential equation of family of parabolas having x axis as line of symmetry and tangent at vertex is y-axes Solution : Let equation of parabola ..........(i) y2 = 4ax dy = 4a .........(ii) 2y dx by (i) and (ii) dy dy ⇒ y = 2x ⇒ y2 = 2yx dx dx Self Practice Problems : (2) Obtain a differential equation of the family of curves y = a sin (bx + c) where a and c being arbitrary constant. (3) Show that the differential equation of the system of parabolas y2 = 4a(x – b) is given by (4) Ans. 2  dy  +   =0 2 dx  dx  Form a differential equation of family of parabolas with focus as origin and axis of symmetry along the x-axis. y d2 y (2) d2 y dx 2 + b2y = 0 (4)  dy  y2 = y2    dx  2 + 2xy dy dx Differential Equation Solution of a Differential Equation: Finding the dependent variable from the differential equation is called solving or integrating it. The solution or the integral of a differential equation is, therefore, a relation between dependent and independent variables (free from derivatives) such that it satisfies the given differential equation NOTE : The solution of the differential equation is also called its primitive, because the differential equation can be regarded as a relation derived from it. There can be three types of solution of a differential equation: (i) General solution (or complete integral or complete primitive) : A relation in x and y satisfying a given differential equation and involving exactly same number of arbitrary constants as order of differential equation. Particular Solution : A solution obtained by assigning values to one or more than one arbitrary (ii) constant of general solution. (iii) Singular Solution : It is not obtainable from general solution. Geometrically, Singular solution acts as an envelope to General solution. 4.1. Differential Equation of First Order and First Degree : dy + f(x, y) = 0, which can also be dx written as : Mdx + Ndy = 0, where M and N are functions of x and y. A differential equation of first order and first degree is of the type Solution methods of First Order and First Degree Differential Equations : 5.1 Variables separable : If the differential equation can be put in the form, f(x) dx = φ(y) dywe say that variables are separable and solution can be obtained by integrating each side separately. A general ∫ ∫ solution of this will be f(x) dx= φ(y) dy + c, where c is an arbitrary constant. Example # 4 : Solve the differential equation (1 + x) y dx = (y – 1) x dy Solution : The equation can be written as  y − 1 1   1+ x  ⇒  x + 1 dx =  x  dx =  y  dy       ∫  1 ∫ 1 − y  dy n x + x = y – ny + c ⇒ ny + nx = y – x + c ⇒ xy = cey–x Example # 5 : Solve : (ex + 1) y dy = (y + 1) ex dx Solution : The given differential equation is (ex + 1) y dy = (y + 1) ex dx ydy ex = (y + 1) (e x + 1) Integrating both sides ⇒ y – log |y + 1|= log (ex + 1) + log k ⇒ y = log |(y + 1)(ex + 1)| + log k ⇒ (y + 1)(ex + 1)=eyc Example # 6 : dy x(2nx + 1) = Solve : dx sin y + y cos y Solution : dy x(2nx + 1) = dx sin y + y cos y (siny + ycosy)dy = x(2nx + 1)dx Integrating both sides  x 2 1 x  x 2 ⇒ y sin y = x2  n x ⇒ – cosy + {(ysiny) + cosy} = 2 ×  ln x − dx  + 2 2 1 2   ∫ 5.1.1 Polar coordinates transformations : Sometimes transformation to the polar co-ordinates facilitates separation of variables. In this connection it is convenient to remember the following differentials: (a) If x = r cos θ ; y = r sin θ then, (iii) x dy – y dx = r2dθ (i) x dx + y dy = r dr (ii) dx2 + dy2 = dr2 (b) If x = r sec θ & y = r tan θ then (i) x dx – y dy = r dr (ii) x dy – y dx = r2 secθ dθ. Differential Equation Example # 7 : Solve the differential equation xdx + ydy = x (xdy – ydx) Solution : Taking x = r cosθ, y = r sinθ x2 + y2 = r2 2x dx + 2ydy = 2rdr xdx + ydy = rdr .........(i) dy −y x y dθ dx = tanθ ⇒ = sec2θ . 2 x dx x xdy – y dx = x2 sec2θ . dθ ........(ii) xdy – ydx = r2 dθ Using (i) & (ii) in the given differential equation then it becomes r dr = r cosθ. r2 dθ dr r 2 ⇒– = cosθ dθ 1 1 = sinθ + λ ⇒ – r x2 + y2 = y x2 + y2 +λ ⇒ y +1 x2 + y2 =c where – λ′ = c ⇒ (y + 1)2 = c(x2 + y2) 5.1.2 Equations Reducible to the Variables Separable form : If a differential equation can be reduced into a variables separable form by a proper substitution, then it is said to be “Reducible to the variables separable type”. Its general form is dy = f(ax + by + c) a, b ≠ 0. To solve dx this, put ax + by + c = t. Example # 8 : Solve Solution : dy = (4x + y + 1)2 dx Putting 4x + y + 1 = t ⇒ 4+ dy dt = ⇒ dx dx dy dt = –4 dx dx Given equation becomes dt – 4 = t2 dx dt ⇒ t +4 2 = dx (Variables are separated) Integrating both sides, dt ∫4+t Example # 9 : Solve Solution : 2 = ∫ dx ⇒ 1 t tan–1 =x+c 2 2 ⇒ 1  4x + y + 1  tan–1   =x+c 2 2    dy  sin–1   =x+y  dx  dy = sin (x + y) dx putting x + y = t dt dt dt dy dt = –1∴ – 1 = sin t ⇒ = 1 + sin t ⇒ = dx 1 + sin t dx dx dx dx Integrating both sides, dt 1 − sin t ∫ 1 + sin t = ∫ dx ⇒ ∫ cos tan t – sec t = x + c ⇒ – 2 t dt = x + c 1 − sin t =x+c cos t substituting the value of t sin (x + y) = x cos (x + y) + c cos (x + y) + 1 ⇒ ⇒ ∫ (sec 2 t − sec t tan t) dt = x + c sin t – 1 = x cos t + c cos t Differential Equation Self Practice Problems : (5) (6) (7) (8) (9) (10) 5.2 dy = (x + 1) (y + 1) dx ydx − xdy xdx + ydy Solve the differential equation = 2 2 x2 x +y Solve the differential equation x2 y dy = ex + y + x2ey dx dy Solve : xy = 1 + x + y + xy dx dy Solve = 1 + ex – y dx dy = sin(x + y) + cos (x + y) dx Solve : dy = 1, satisfying the condition y(1) = 0 dx y =c x2 + y2 + x (11) Find the solution of the differential equation (x + y)2 Ans. (5) y – n (y + 1) = nx – (7) – (9) ey – x = x + c (11) y+ 1 e y = ex + 1 +c x x3 +c 3 (6) (8) y = x + n |x (1 + y)| + c (10) log tan x+y +1 2 =x+c π = tan–1(x + y) 4 Homogeneous Differential Equations : dy f(x,y) = where f and g are homogeneous function of dx g(x,y) x and y, and of the same degree, is called homogeneous differential equaiton and can be solved easily by putting y = vx. A differential equation of the form Example # 10 : Solve x Solution : dy y = y + x tan   dx x y y dy = + tan (Homogeneous differential equation) x x dx dv put y = vx ⇒ v+x = v + tan v dx Integrating both sides we have ⇒ n sin v = nx + nc sinv = cx ⇒ ⇒ cot v.dv = dx x y sin   = cx x Example # 11 : Solve : x2dy + y (x + y)dx = 0 given that y = 1 when x = 1 y y2 dy Solution : =– − 2 put y = vx x x dx dx dv dy dv dv =v+x ⇒ v+x = – v – v2 ⇒ =– x (2 + v)v dx dx dx 1 1  ⇒ 1 2 ∫  v – v + 2  ⇒ x v =C v+2 dv = When x = 1 then y = 1 ⇒ C = ∫– 1 3 dx x ⇒ 1  v  n + nx = nC 2  v + 2  ⇒ x ⇒ 3x2y = (y + 2x) y =C y + 2x Differential Equation 5.2.1 Equations Reducible to the Homogeneous form dy ax + by + c = .........(1) dx Ax + By + C can be made homogeneous (in new variables X and Y) by substituting x = X + h and y = Y + k, where h aX + bY + (ah + bk + c) dY = ........(2) and k are constants to obtain, dX AX + BY + (Ah + Bk + C) These constants are chosen such that ah + bk + c = 0, and Ah + Bk + C = 0. Thus we obtain the aX + bY dY = following differential equation dX AX + BY The differential equation can now be solved by substituting Y = vX. Equations of the form x + 2y − 5 dy = dx 2x + y − 4 y=Y+k Example # 12 : Solve the differential equation Solution : Let x = X + h, dy d = (Y + k) dX dX dy dY dx = ...........(i) =1+0 ...........(ii) dX dX dX dy dY on dividing (i) by (ii) = dx dX dY X + h + 2(Y + k) − 5 X + 2Y + (h + 2k − 5) = = dX 2X + 2h + Y + k − 4 2X + Y + (2h + k − 4) h & k are such that h + 2k – 5 = 0 & 2h + k – 4 = 0 h = 1, k = 2 dY X + 2Y = which is homogeneous differential equation. dX 2X + Y Now, substituting Y = vX 1 + 2v dX dv dY dv 2+v =v+X ∴X = –v ⇒ dv = 2+v X dX dX dX 1− v2  1 3 1 3  ⇒ n (v + 1) – n (1 – v) = n X + c +   dv = n X + c 2 2 2(v + 1) 2(1 − v)   ∫ ∫ ∫ n v +1 (1 − v)3 = nX2 + 2c ⇒ (Y + X) X2 (X − Y)3 X2 = e2c where e2c = c′ X + Y = c′(X – Y)3 x – 1 + y – 2 = c′ (x – 1– y + 2)3 x + y – 3 = c′ (x – y + 1)3 Special case : Case - 1 In equation (1) if a b , then the substitution ax + by = v will reduce it to the form in which variables = A B are separable. 2x + 3y − 1 dy = 4x + 6y − 5 dx Putting u = 2x + 3y du dy =2+3. ⇒ dx dx Example # 13 : Solve Solution : 1  du u −1 du 3u − 3 + 4u − 10  ⇒ = − 2 = 3  dx dx 2u − 5 2u − 5  2 9 2u − 5 1 dx = dx ⇒ . du = x + c 1.du – 7 7 7u − 13 7u − 13 2 9 1 9 ⇒ u– . n (7u – 13) = x + c ⇒ 4x + 6y – n (14x + 21y – 13) = 7x + 7c 7 7 7 7 9 ⇒ – 3x + 6y – n (14x + 21y – 13) = c′ 7 In equation (1), if b + A = 0, then by a simple cross multiplication equation (1) becomes an exact differential equation. ∫ Case - 2 ∫ ∫ ∫ Differential Equation x − 2y + 5 dy = 2x + y − 1 dx Cross multiplying, 2xdy + y dy – dy = xdx – 2ydx + 5dx 2 (xdy + y dx) + ydy – dy = xdx + 5 dx 2 d(xy) + y dy – dy = xdx + 5dx On integrating, y2 x2 –y= + 5x + c ⇒ x2 – 4xy – y2 + 10x + 2y = c′ 2xy + 2 2 Example # 14 : Solve Solution : Case - 3 where c′ = – 2c If the homogeneous equation is of the form : yf(xy) dx + xg(xy)dy = 0, the variables can be separated by the substitution xy = v. Self Practice Problems : Solve the following differential equations (12) y  dy  –1  x dx − y  tan x = x given that y = 0 at x = 1 (13)   x (14) dy 2x – y + 3 = dx x + 2y + 4 (15) (3x – 2y + 1) dy + (4y – 6x + 3)dx = 0 (16) dy 3x + 2y − 5 = dx 3y − 2x + 5 (13) x sin Ans. 5.3 x +y 2 (12) 2 = y y tan−1 x x e (14) y2 – x2 + xy + 4y – 3x + C = 0 (16) 3x + 4xy – 3y – 10x – 10y = C 2 (15) y dy = y – x tan x dx y =C x 10 n |3x – 2y – 9| = 2y – 4x + C 2 Exact Differential Equation : dy =0 ...........(1) dx Where M and N are functions of x and y is said to be exact if it can be derived by direct differentiation (without any subsequent multiplication, elimination etc.) of an equation of the form f(x, y) = c dx = 0 is an exact differential equation. e.g. y2 dy + x dx + x The differential equation M + N NOTE : (i) (ii) ∂M ∂N . = ∂y ∂x For finding the solution of exact differential equation, following results on exact differentials should be remembered: xdy − ydx y = d  (a) xdy + y dx = d(xy) (b) 2 x x The necessary condition for (1) to be exact is (c) (e) (g) 2(x dx + y dy) = d (x2 + y2) xdy − ydx x +y 2 2 xdy + ydx 2 x y 2 y  = d  tan−1  x   1  = d−   xy  Example # 15 : Solve : y2x dx + ydx – xdy = 0 Solution : y 2 xdx + ydx – xdy y 2 on integrating =0 x ⇒ xdx + d   = 0 y x2 x + +c=0 2 y (d) xdy − ydx  y = d  ln  xy  x (f) xdy + ydx = d(ln xy) xy Differential Equation Example # 16 : Solve : (x – y)dy + (x + y)dx = dx + dy Solution : The given equation can be written as (xdy + ydx) – ydy + xdx = dx + dy ⇒ d(x.y) – ydy + xdx = dx + dy 2 2 y x Also integrating each term we get xy – =x+y+C + 2 2 Self Practice Problems : (17) Solve : xdy + ydx + xy ey dy = 0 (18) Solve : ye–x/y dx – (xe–x/y + y3) dy = 0 Ans. (17) n (xy ) + ey = c (18) 2e–x/y + y2 = c Linear Differential Equation : A linear differential equation has the following characteristics : (i) Dependent variable and its derivative in first degree only and are not multiplied together (ii) All the derivatives should be in a polynomial form (iii) The order may be more than one The mth order linear differential equation is of the form. dm y dm−1y dy + .................... + Pm–1 (x) + Pm (x) y = φ(x), dx dx dxm−1 whereP0(x), P1(x)...............Pm(x) are called the coefficients of the differential equation. P0(x) NOTE : 6.1 m + P1(x) dy + y2 sinx = lnx is not a Linear differential equation. dx Linear differential equations of first order : dy + Py = Q , is linear in y. dx (Where P and Q are functions of x only). The differential equation Integrating Factor (I.F.) : It is an expression which when multiplied to a differential equation converts it into an exact form. I.F for linear differential equation = e ∫ Pdx (constant of integration will not be considered) ∴ after multiplying above equation by Ι.F it becomes; dy ∫ Pdx + Py . e ∫ Pdx = Q. e ∫ Pdx e dx d ⇒ ⇒ y.e ∫ Pdx = Q.e ∫ Pdx + C . (y.e ∫ Pdx ) = Q.e ∫ Pdx dx ∫ NOTE : Some times differential equation becomes linear, if x is taken as the dependent variable, and y dx + P1 x = Q1. as independent variable. The differential equation has then the following form : dy where P1 and Q1 are functions of y. The Ι.F. now is e ∫ dy Example # 17 : Solve the differential equation f(x) = f 2(x) + f(x) y + f′(x) y dx dy Solution : f(x) = f 2(x) + f(x) y + f′(x) y dx f ′(x)  dy  – 1 + Given DE can be written as  y = f(x) f(x)  dx  Which is L.D.E. P1 dy I.F. = e–x -n f(x) = e– x f(x) General solution y e– x = f(x) ∫ f(x) e– x dx + c = – e–x + c ⇒ y = – f(x) + cex f(x) f(x) Differential Equation dy + y = 2 n x dx 1 2 dy + y= ⇒ xnx x dx Example # 18 : Solve : x nx Solution : IF = e ∫ ∴ P.dx P= 1 2 ,Q= xnx x 1 =e ∫ xnx dx = en( nx) = n x 2 ∫ x .nx.dx + c General solution is y. (n x) = ⇒ y (n x) = (n x)2 + c Example # 19 : Solve the differential equation t (1 + t2) dx = (x + xt2 – t2) dt and it given that x = – π/4 at t = 1 Solution : t (1 + t2) dx = [x (1 + t2) – t2] dt x t dx = − t (1 + t 2 ) dt dx x t − = − dt t 1 + t2 which is linear in t 1 , Q=– t 1 + t2 ∴ General solution is 1 t  1 = dt + c x: . − t  1 + t 2  t putting x = – π/4, t = 1 – π/4 = – π/4 + c ⇒ c=0 ∴ x = – t tan–1 t Here, P = – IF = e ∫ 6.2 6.2.1 − ⇒ dx dt 1 ∫ t dt = e–nt = 1 t x = – tan–1 t + c t Equations reducible to linear form By change of variable. Often differential equation can be reduced to linear form by appropriate substitution of the non-linear term dy = cos x (sinx – y2) dx The given differential equation can be reduced to linear form by change of variable by a suitable subtitution. Substituting y2 = z dy dz = 2y dx dx differential equation becomes sin x dz + cos x.z = sin x cos x 2 dx dz dz + 2 cot x . z = 2 cos x which is linear in dx dx Example # 20 : Solve : y sinx Solution : IF = e ∫ = e2n sin x = sin2 x General solution is 2 cot x dx ∴ z. sin2 x = 6.2.2 ∫ 2cos x.sin 2 x.dx + c ⇒ y2 sin2x = 2 sin3 x + c 3 Bernoulli’s equation : dy + Py = Q.yn, n ≠ 0 and n ≠ 1 dx where P and Q are functions of x, is called Bernoulli’s equation and can be made linear in v by dividing by yn and putting y –n+1 = v. Now its solution can be obtained as in (v). dy – y cos x = xy3 ex . e.g. : 2 sin x dx Equations of the form Differential Equation dy = x3y3 – xy (Bernoulli's equation) dx Dividing both sides by y3 1 dy x + 2 = x3 3 dx y y Example # 21 : Solve : Solution : Putting t = 1/y2 ⇒ – dt 2 dy = 3 dx dx y ⇒ dt – 2x t = –2x3 dx Ι.F. = e− ∫ 2xdx = e− x General solution is 2 te – x2 = ∫ –2x e 3 – x2 dx + C ⇒ e– x y 2 1 2 = – e – x (–x2 – 1) + C ⇒ 2 y 2 = (x2 + 1) + C e x 2 Self Practice Problems : (19) (20) (21) dy = y (1 – x2) + x2 nx dx dy Solve : (x + 2y3) =y dx dy Solve : x + y = y2 log x dx Solve : x (x2 + 1) (22) Solve the differential equation Ans. (19) (21)  dy  3 3 xy2   – 2y = 2x given y = 1 at x = 1  dx   x2 + 1    y = x n x – x + c  x  y (1 + cx + log x) = 1 (20) x = y (c + y2) (22) y3 + 2x3 = 3x6 Higher Degree Equation : The differential equation y = mx + f(m) (where m= dy ) dx ............(1), is known as Clairaut’s Equation. To solve (1), differentiate it w.r.t. x, which gives d f(m) dm dy dm =m+x + dm dx dx dx d f(m) dm dm + =0 dm dx dx dm either =0⇒m=c ...........(2) or x + f′(m) = 0 ............(3) dx NOTE : (i)If m is eliminated between (1)and(2),thesolutionobtained is a general solution of(1) (ii) If m is eliminated between (1) and (3), then solution obtained does not contain any arbitrary constant and is not particular solution of (1). This solution is called singular solution of (1). dy Example # 22 : Solve : y = mx + m – m3 where, m = dx Solution : y = mx + m – m3 ..... (i) The given equation is in clairaut's form. Now, differentiating wrt. x dm dm dm dy dm dm dm ⇒ m=m+x + − 3m2 =m + x + − 3m2 dx dx dx dx dx dx dx dm dm (x + 1 – 3m2) = 0 ⇒ =0⇒ m=c ..... (ii) dx dx x +1 or x + 1 – 3m2 = 0 ⇒ m2 = ..... (iii) 3 Eliminating 'm' between (i) & (ii) is called the general solution of the given equation. y = cx + c – c3 where, 'c' is an arbitrary constant. Again, eliminating 'm' between (i) & (iii) is called singular solution of the given equation. y = m (x + 1 – m2) x Differential Equation 1/ 2  x + 1 y = ±   3   x + 1 y = ± 2   3  Self Practice Problems : x + 1   x + 1− 3    1/ 2 ⇒  x + 1 y = ±   3  ⇒ y2 = 3/2 4 (x + 1)3 ⇒ 27y2 = 4 (x + 1)3 27 (23) Solve the differential equation y = mx + 2/m where, m = (24) Solve : sin px cos y = cos px sin y + p where p = Ans. (23) (24) 2 (x + 1) 3 dy dx dy dx General solution : y = cx + 2/c where c is an arbitrary constant Singular solution : y2 = 8x General solution : y = cx – sin–1 (c) where c is an arbitrary constant. Singular solution : y = x 2 − 1 − sin−1 x2 − 1 x2 Example # 23 : The normal to a given curve at each point (x,y) on the curve passes through the point (3,0). If the curve contains, the point (3,4) find its equation. Solution : Equation of normal at any point (x,y) is dy (Y – y) + (X – x) = 0 dx Passes through (3,0) y2 x2 dy – y = 0 ⇒ ydy = (3 – x)dx ⇒ = 3x – 2 2 dx The curve contains the point (3,4) 9 ⇒ 8 = 9 – +C ⇒ C = 7/2 2 By equation (i) ⇒ (3 – x) +C .......... (i) y2 x2 7 = 3x – + ⇒ x2 + y2 – 6x – 7 =0 2 2 2 y y y – cot . cos , x x x (x > 0, y > 0) and curve passes through the point(1,π/4). Find the equation of the curve. Let y = f(x) be the curve y y y dy = – cot . cos (homogeneous differential equation) given that x x x dx put y = vx –dx dv = v – cotv. cosv ⇒ tan v. sec v dv = ⇒ v+x x dx Integrating both sides, we have y + n|x| = C ⇒ secv = – n|x| + C ⇒ sec x Example # 24 : The slope of the tangent to a curve at any point (x,y) on it given by Solution :  π Passes through  1,  ⇒ C = 2  4 The curve is ⇒ sec y +n|x| = 2 x Differential Equation Example # 25 : Assume that a spherical rain drop evaporates at a rate proportional to its surface area. if its radius originally is 3mm and 1 hr later has been reduced to 2mm, find an expression for radius of the rain drop at any time. Solution : Let r be radius , V be volume and S be surface area of rain drop at any time t. Then V = given ⇒ 4 3 πr and 3 dV ∝S ⇒ dt 4 dr .3πr2 = k4πr2 3 dt S = 4 πr2 dV = kS, k is constant of proportionality dt ⇒ dr =k dt ⇒ dr = kdt Integrating both sides we have r = kt + C when t = 0, r = 3 ⇒ C = 3 when t = 1 hr, r = 2 ⇒ k = – 1 Hence r = 3 – t Ans. Self Practice Problems : (25) The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass. Ans. (25) k log 2 , where k is constant of proportionality