Differential Equations

Rule for solving homogeneous equation Let the given equation be homogeneous. Then, by definition, the given equation can be put in the form dy/dx = ƒ(y/x) ---------- (1) Solution Method: - To solve (1), let y/x = v or, y = vx Differentiating equation (1) with respect to x, we get dy/dx = v + x (dv/dx)----------- (2) So the equation becomes v + x (dv/dx) = ƒ (v) or, x (dv/dx) = ƒ (v) – v Separating the variables x and v, we have dx/x = dv/{ƒ (v) – v}--------------(3) Integrating equation (3) After integration, replace v by y/x and finally we will find the required solution. Example: Solve: (x 3 + 3xy 2 )dx + (y 3 + 3x 2 y)dy = 0 Given that, (x 3 + 3xy 2 )dx + (y 3 + 3x 2 y)dy = 0 or, dy/dx = - {(x 3 + 3xy 2 )/ (y 3 + 3x 2 y)} or, dy/dx = - [{1 + 3(y/x) 2 }/{(y/x) 3 + 3(y/x)}] -------(1) Take, y/x = v or, y = vx Differentiating with respect to x, we get dy/dx = v + x (dv/dx)----------- (2) From equation (1) and (2) v + x (dv/dx)

Transformation of some equation in the form in which variables are separable . Equation of the form Dy/dx = ƒ (ax + by + c) or, dy/dx = ƒ (ax + by) can be reduced to an equation in which variables can be separated. For this purpose we use the substitution ax + by + c = v or, ax + by = v Example: Solve: dy/dx = (4x + y +1) 2 Given that, dy/dx = (4x + y +1) 2 ----------- (1) Let, 4x + y +1 = v -------------- (2) Differentiating equation (1) with respect to x, we get 4 + dy/dx = dv/dx or, dy/dx = dv/dx – 4 --------------- (3) We get from equation (2) & (3), dv/dx – 4 = v 2 or, dv/dx = 4 + v 2 or, dx = dv/(4 + v 2 ) Integrating, x + c = (1/2) tan -1 (v/2) or, 2x + 2c = tan -1 (v/2) or, v = 2 tan (2x + 2c) or, 4x + y +1 = 2 tan (2x + 2c) Answer: Example: Solve: dy/dx = sin (x + y) + cos (x + y) Given that, dy/dx = sin (x + y) + cos (x + y) ------------- (1) Let, x + y = v ------------(2) Differentiating equation (1) with

Separation of variables If a differential equation of the first order and first degree is of the form ƒ 1 (x)dx = ƒ 2 (y)dy Solve: y –x dy/dx = a(y 2 + dy/dx) Solution: - Given that, y –x dy/dx = a(y 2 + dy/dx) => y – ay 2 = dy/dx (a + x) => y (1 - ay)/dy = (a + x)/dx => dx/(a + x) = dy/y (1 - ay) => dx/(a + x) = [{a/(1 - ay)} + 1/y] dy Integrating, ln(a + x) = - ln(1 - ay) + lny + lnc => ln(a + x) = ln{cy/(1 - ay)} => a + x = cy/(1 - ay) Answer: - Solve: dy/dx = x (2 lnx + 1)/siny +y cosy Solution: - Given that, dy/dx = x (2 lnx + 1)/siny +y cosy =>siny +y cosy)dy = (2xlnx + x)dx Integrating, ſ sinydy + ſ y cosydy = 2 ſ x lnxdx + ſxdx => - cosy + y ſ cosydy - ſ (dy/dy ſ cosydy)dy = 2 lnx ſ xdx - 2 ſ(d lnx/dx ſ xdx)dx + x 2 /2 => - cosy +y siny - ſ sinydy = x 2 lnx - ſ xdx + x 2 /2 => - cosy + y siny + cosy = x 2 lnx - x 2 /2+ x 2 /2 + c => y siny = x 2 lnx + c Answer: -