Differential Equations

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: -

a)Find the differential equation of the family of curves y = e x (A cosx + B sinx), where A and B are arbitrary constants. Solution: - Given that, y = e x (A cosx + B sinx) ------------ (1) Differentiating (1) with respect to x, we get y ‘ = e x (-A sinx + B cosx) + e x (A cosx + B sinx) or, y ‘ = e x (-A sinx + B cosx) + y {using (1)} ------ (2) Differentiating (2) with respect to x, we get y “ = -e x (A cosx + B sinx) + e x (-A sinx + B cosx) + y’ ---- (3) Now From Equation (1) e x (-A sinx + B cosx) = y’ – y -----------(4) Hence eliminating A and B from equation (1), (3) and (4), we get y” = - y + y’ – y + y’ or, y” – 2y’ +2y = 0 ---------- (5) Equation (5) is the required differential equation. b) By eliminating the constants a and b obtain the differential equation for which xy = ae x + be -x + x 2 is a solution. Solution: - Given that, xy = ae x + be -x + x 2 ---------- (1) Differentiating (1) with respect to x,