Homogeneous First-Order Differential Equations

Check whether the expression depends on y/x or x/y, or whether numerator and denominator have the same degree. Choose y = vx or x = vy. Differentiate using the product rule. Substitute into the differential equation. Separate v and x, integrate, add C, and return to x and y.

A diagram is not required for standard CBSE algebraic solving. Homogeneity here means degree pattern in x and y, not visual symmetry of a graph.

Treating y = vx as if v were constant, which gives dy/dx = v and destroys the equation. Another common error is not replacing v by y/x in the final answer.

A first-order differential equation dy/dx = F(x, y) is homogeneous when the right side can be expressed as a function of y/x or x/y. It is solved by substituting y = vx or x = vy to reduce it to a variable-separable equation.

A first-order differential equation dy/dx = F(x, y) is homogeneous when the right side can be expressed as a function of y/x or x/y. It is solved by substituting y = vx or x = vy to reduce it to a variable-separable equation. Condition: dy/dx = F(y/x) or dy/dx = F(x/y), or M(x, y)dx + N(x, y)dy = 0 with M and N homogeneous of the same degree. Main substitution: y = vx, so dy/dx = v + x(dv/dx). Then solve the resulting separable equation and replace v by y/x. For dy/dx = (x + y)/x, write y/x = v. Then dy/dx = 1 + v and also dy/dx = v + x(dv/dx). Hence v + x(dv/dx) = 1 + v, so x(dv/dx) = 1, giving v = log|x| + C and y/x = log|x| + C.

In equations of the form dy/dx = f(x, y), homogeneity is checked by seeing whether f(tx, ty) = f(x, y), or whether numerator and denominator are homogeneous expressions of the same degree. With y = vx, y changes with x, so dy/dx = v + x(dv/dx). This converts the equation into one involving v and x, which is usually separable. Condition: dy/dx = F(y/x) or dy/dx = F(x/y), or M(x, y)dx + N(x, y)dy = 0 with M and N homogeneous of the same degree. Main substitution: y = vx, so dy/dx = v + x(dv/dx). Then solve the resulting separable equation and replace v by y/x. Example: Solve dy/dx = (x + y)/x. Put y = vx, so dy/dx = v + x(dv/dx). Then v + x(dv/dx) = 1 + v. Thus x(dv/dx) = 1, so dv = dx/x. Integrate: v = log|x| + C. Replace v: y/x = log|x| + C, hence y = x(log|x| + C). Asked as a solving question where the equation is not separable at first glance. Students are expected to recognize equal-degree expressions and use y = vx or x = vy correctly. Treating y = vx as if v were constant, which gives dy/dx = v and destroys the equation. Another common error is not replacing v by y/x in the final answer.