Show that (x^2+xy)dy=(x^2+y^2)dx is homogeneous and solve it.
Hint. Put y=vx to turn the right-hand side into a function of v alone, then separate variables.
With y=vx, dy/dx=v+x(dv/dx), and the equation becomes v+x(dv/dx)=(1+v^2)/(1+v); simplifying the right side minus v gives x(dv/dx)=(1-v)/(1+v), a separable equation in v and x.
✦ log|x/(x-y)^2|-y/x=C
Show that y'=(x+y)/x is homogeneous and solve it.
Hint. Put y=vx — the right-hand side becomes 1+v directly.
With y=vx, dy/dx=v+x(dv/dx), the equation becomes v+x(dv/dx)=1+v, so x(dv/dx)=1, a directly separable equation.
✦ y=x(log|x|+C)
Show that (x-y)dy-(x+y)dx=0 is homogeneous and solve it.
Hint. Put y=vx, then split the resulting fraction into a tan^-1 piece and a log piece.
With y=vx, the equation reduces to x(dv/dx)=(1+v^2)/(1-v), and splitting (1-v)/(1+v^2) into 1/(1+v^2) minus v/(1+v^2) gives a tan^-1 term and a log term on integration.
✦ tan^-1(y/x)=(1/2)log(x^2+y^2)+C
Show that (x^2-y^2)dx+2xy dy=0 is homogeneous and solve it.
Hint. Put y=vx — the resulting separable equation in v integrates to a simple log.
With y=vx, the equation reduces to 2v/(1+v^2) dv=-dx/x, and since the left side integrates to log(1+v^2), combining with the right side gives a compact relation between x and y.
✦ x^2+y^2=Cx
Show that x^2(dy/dx)=x^2-2y^2+xy is homogeneous and solve it.
Hint. Put y=vx — the resulting equation separates into dv/(1-2v^2) on one side.
Since substituting y=vx reduces the equation to x(dv/dx)=1-2v^2, a separable equation whose left side is a standard 1/(a^2-b^2v^2) form, integrating gives a log-ratio.
✦ (1/(2*sqrt(2))) log|(x+sqrt(2)y)/(x-sqrt(2)y)| = log|x|+C
Show that x dy-y dx=sqrt(x^2+y^2) dx is homogeneous and solve it.
Hint. Put y=vx — the resulting equation is the standard dv/sqrt(1+v^2) integral, a sinh^-1 form.
With y=vx, the equation reduces to x(dv/dx)=sqrt(1+v^2), a separable equation whose left side integrates to log|v+sqrt(1+v^2)|, giving a clean relation once simplified.
✦ y+sqrt(x^2+y^2)=Cx^2
Show that {x cos(y/x)+y sin(y/x)}y dx={y sin(y/x)-x cos(y/x)}x dy is homogeneous and solve it.
Hint. Put y=vx — the trigonometric terms combine into a single tan(v)/2 minus 1/(2v) integral.
Putting y=vx reduces the equation to x(dv/dx)=2v cos(v)/(v sin(v)-cos(v)); since separating and splitting into tan(v)/2 minus 1/(2v) integrates to a log of v cos(v), this gives a compact xy relation.
✦ xy cos(y/x)=C
Show that x(dy/dx)-y+x sin(y/x)=0 is homogeneous and solve it.
Hint. Put y=vx — the resulting equation separates with cosec(v) on the left.
With y=vx, the equation reduces to x(dv/dx)=-sin(v), a separable equation whose left side integrates via cosec(v) to a log of tan(v/2).
✦ y=2x.tan^-1(C/x) (equivalently tan(y/(2x))=C/x)
Show that y dx+x log(y/x) dy-2x dy=0 is homogeneous and solve it.
Hint. This is more naturally homogeneous in x as a function of y — put x=vy.
Treating x as the dependent variable and putting x=vy reduces the equation to a separable equation in v=x/y; since the left side integrates via w=1-log(v) to a log of w, this gives a compact relation.
✦ y[1-log(x/y)]=C, i.e. y[1+log(y/x)]=C
Show that (1+e^(x/y))dx+e^(x/y)(1-x/y)dy=0 is homogeneous and solve it.
Hint. This is more naturally homogeneous in x as a function of y — put x=vy.
With x=vy, treating x as the dependent variable, the equation reduces to (1+e^v)/(v+e^v) dv=-dy/y; since the left side's numerator is exactly the derivative of its denominator, it integrates directly to a log.
✦ x+y.e^(x/y)=C
Find a particular solution: (x+y)dy+(x-y)dx=0; y=1 when x=1.
Hint. This is the homogeneous equation dy/dx=(y-x)/(x+y) — put y=vx and separate.
Putting y=vx reduces the equation to a separable form whose left side (after splitting into a tan^-1 piece and a log piece) integrates to tan^-1(y/x) plus half the log of x^2+y^2; since y(1)=1 is given, applying it fixes the constant.
✦ tan^-1(y/x)+(1/2)log(x^2+y^2)=pi/4+(1/2)log2
Find a particular solution: x^2 dy+(xy+y^2)dx=0; y=1 when x=1.
Hint. This is homogeneous — put y=vx and separate the variables.
Because substituting y=vx reduces the equation to a separable form in v and x, integrating and substituting back v=y/x, then applying y(1)=1, fixes the constant.
✦ y=2x/(3x^2-1)
Find a particular solution: [x sin^2(y/x)-y]dx+x dy=0; y=pi/4 when x=1.
Hint. This is homogeneous — put y=vx, which reduces the equation to a separable one in cot(v).
This equation becomes separable once y=vx is substituted, reducing it to x(dv/dx)=-sin^2(v); since the left side integrates to -cot(v), applying y(1)=pi/4 fixes the constant.
✦ cot(y/x)=log|x|+1
Find a particular solution: (dy/dx)-(y/x)+cosec(y/x)=0; y=0 when x=1.
Hint. This is homogeneous — put y=vx, which reduces the equation to a separable one in cos(v).
Substituting y=vx here turns the equation into the separable form x(dv/dx)=-cosec(v); because the left side integrates to -cos(v), applying y(1)=0 fixes the constant.
✦ y=x.cos^-1(log|x|+1)
Find a particular solution: 2xy+y^2-2x^2(dy/dx)=0; y=2 when x=1.
Hint. This is homogeneous — put y=vx, which reduces the equation to a separable one where v^2/2 appears on one side.
With y=vx, the equation reduces to x(dv/dx)=v^2/2, a separable equation whose left side integrates to -1/v; applying y(1)=2 fixes the constant.
✦ y=2x/(1-log|x|)
Choose the correct answer: a homogeneous differential equation of the form dx/dy=h(x/y) can be solved by making the substitution (A) y=vx (B) v=yx (C) x=vy (D) x=v
Hint. Recall which substitution matches the given form — the independent variable here is y, and the dependent variable is x.
Since the equation is given as dx/dy=h(x/y), with x as the dependent variable, the substitution treats the ratio x/y as v, i.e. x=vy — the mirror image of the more familiar y=vx substitution used when dy/dx=h(y/x).
✦ (C) x=vy
Choose the correct answer: which of the following is a homogeneous differential equation? (A) (4x+6y+5)dy-(3y+2x+4)dx=0 (B) (xy)dx-(x^3+y^3)dy=0 (C) (x^3+2y^2)dx+2xy dy=0 (D) y^2 dx+(x^2-xy-y^2)dy=0
Hint. Check whether every term in the equation has the same total degree in x and y — that is the test for homogeneity.
In option (D), every term (y^2, x^2, xy, y^2) has total degree exactly 2 in x and y, so F(lambda x,lambda y)=lambda^2 F(x,y) holds; the other options mix degrees (constants added to degree-1 terms, or degree 2 mixed with degree 3), failing the homogeneity test.
✦ (D) y^2 dx+(x^2-xy-y^2)dy=0
Solutions written by the tuition.in editorial team and checked against lemh203.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.
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