Q.Find the particular solution of the differential equation log(dxdy)=3x+4y given that y=0 when x=0.
Concept understanding — Separation Of Variables
Separation of Variables: From Intuition to Precision
Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.
The Core Intuition
Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:
- The temperature difference between the coffee and the room (a function of time)
- The surface area of the cup (a function of shape, not time)
These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.
The Precise Statement
Separation of Variables applies to ordinary differential equations (ODEs) of the form:
dxdy=f(x)⋅g(y)
where the right-hand side is a product of a function of x alone and a function of y alone. The method works in three clean steps:
dxdy=f(x)⋅g(y)⟹g(y)1dy=f(x)dx
Step 1: Separate. Multiply both sides by dx and divide by g(y) (assuming g(y)=0). This moves all y's to one side and all x's to the other.
Step 2: Integrate. Put an integral sign on both sides:
∫g(y)1dy=∫f(x)dx
Step 3: Solve. Evaluate both integrals and solve for y explicitly if possible.
You cannot separate if the equation is not in product form. For example, dxdy=x+y cannot be separated — the sum x+y is not a product f(x)g(y).
Why This Works
The justification is the chain rule in reverse. From dxdy=f(x)g(y), rewrite it as:
g(y)1dxdy=f(x)
Now integrate both sides with respect to x:
∫g(y)1dxdydx=∫f(x)dx
The left side is a substitution waiting to happen: dxdydx=dy, so you get ∫g(y)1dy. That's the entire trick — the chain rule dressed up.
A Concrete Example
Solve dxdy=2xy, with y(0)=3.
Step 1: Separate. Divide both sides by y (assuming y=0):
y1dy=2xdx
Step 2: Integrate.
∫y1dy=∫2xdx⟹log∣y∣=x2+C
Step 3: Solve for y.
∣y∣=ex2+C=eC⋅ex2
Let A=±eC (absorbing the absolute value): y=Aex2. Now use y(0)=3: 3=Ae0=A, so A=3.
Final answer: y=3ex2
Always check if g(y)=0 gives a solution. Here g(y)=y, so y=0 is also a solution (the trivial one). Our initial condition y(0)=3 picks the non-zero branch.
When Does It Apply?
Separation of Variables works for first-order ODEs of the form dxdy=f(x)g(y), and for certain partial differential equations like the heat equation (a more advanced use — the idea is the same: assume the solution is a product of functions of each variable). It does not work for equations where x and y are added, subtracted, or composed in non-product ways, or for higher-order ODEs (generally).
The Big Picture
Separation of Variables is your first real tool for solving differential equations. It reduces a problem with two moving parts into two independent single-variable integrals: you separate the variables, handle each alone, then reassemble with the initial condition. Think of it as untangling a knot by pulling the two ends apart — once separate, each piece is easy to deal with.
Separation of Variables is the very first solving technique taught in NCERT's Class 12 Differential Equations chapter and one of the most heavily tested skills in CBSE board exams and JEE Main. Students searching "separable differential equations important questions" or "differential equations class 12 formula" will find this method the starting point for almost every other technique in the chapter.
Remove the log by exponentiating: dxdy=e3x+4y=e3xe4y, which separates:
e−4ydy=e3xdx.
Integrate:
−41e−4y=31e3x+C.
Apply y=0 at x=0: −41=31+C⇒C=−127.
Multiply through by −12:
3e−4y=7−4e3x.
4e3x+3e−4y=7.
Exponentiating turns the equation into a separable one; with y(0)=0 the particular solution is 4e3x+3e−4y=7.
Remove the logarithm
log(dxdy)=3x+4y ⇒ dxdy=e3x+4y=e3xe4y.
Now the right side is a product of a function of x and a function of y, so it separates.
Separate
Divide by e4y (never zero) and multiply by dx:
e−4ydy=e3xdx.
Integrate
∫e−4ydy=∫e3xdx ⇒ −41e−4y=31e3x+C.
Apply the initial condition
At x=0, y=0:
−41e0=31e0+C ⇒ −41=31+C ⇒ C=−127.
Clean up
Multiply −41e−4y=31e3x−127 by −12:
3e−4y=−4e3x+7 ⇒ 4e3x+3e−4y=7.
Check: differentiating gives 12e3x−12e−4yy′=0, so y′=e3xe4y and logy′=3x+4y; also 4+3=7 at the origin. ✓
4e3x+3e−4y=7.
Method: Remove a logarithm first, then separate variables
Use this when the derivative is trapped inside a function — most commonly log(dxdy)=(⋯). Undo the outer function before attempting to separate.
Steps
Step 1: Invert the outer function to free dxdy.
From log(dxdy)=3x+4y, exponentiate: dxdy=e3x+4y.
Step 2: Split the exponential into a product.
Use e3x+4y=e3xe4y so the right side is a function of x times a function of y — now separable.
Step 3: Separate and integrate.
e−4ydy=e3xdx,∫e−4ydy=∫e3xdx.
Step 4: Apply any initial condition and tidy.
Substitute the data point to find the constant, then clear fractions to a neat implicit form.
Common Mistakes
Mistake 1: Trying to separate before removing the logarithm.
Why it's wrong: with dxdy locked inside log, the variables cannot be separated as written. Correct approach: exponentiate first to get dxdy=e3x+4y.
Mistake 2: Not splitting e3x+4y into a product.
Why it's wrong: separation needs e3x+4y=e3xe4y; leaving it combined hides the separable structure. Correct approach: use the index law to split, then write e−4ydy=e3xdx.
Mistake 3: Integration or sign errors with the exponentials.
Why it's wrong: ∫e−4ydy=−41e−4y; a missing −41 or sign error spoils the constant. Correct approach: integrate carefully, then apply x=0,y=0 to fix the constant.
- TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQQ.The general solution of the differential equation x2dy−(xy−y2)dx=0 is (A) y2=3x2log(cx) (B) y2=logx+c (C) ylogx=x+cy (D) ylogx=x2+c
›Reveal solutionSolution
The equation is homogeneous; the substitution y=vx integrates to ylogx=x+cy — option (C).
Rewrite the equation x2dy−(xy−y2)dx=0 as
dxdy=x2xy−y2=xy−(xy)2.
This is homogeneous. Put y=vx, so dxdy=v+xdxdv:
v+xdxdv=v−v2⇒xdxdv=−v2.
Separate variables:
v2dv=−xdx⇒−v1=−logx+c1.
Since v1=yx:
yx=logx+C.
Multiply through by y:
x=ylogx+Cy⇒ylogx=x−Cy=x+cy,
writing c=−C for the arbitrary constant. This matches option (C).
✓Final answerThe general solution is option (C): ylogx=x+cy.
- TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQQ.The general solution of the differential equation dxdy=xy−2x+y−2xy+x−2y−2 is (A) x+y+3logy+1x+1=c (B) x+y+3logx+1y+1=c (C) x−y+3logx+1y+1=c (D) x−y+3logy+1x+1=c
›Reveal solutionSolution
The equation separates into (y−2)/(y+1)dy = (x−2)/(x+1)dx, integrating to x − y + 3log|(y+1)/(x+1)| = c.
Numerator xy+x−2y−2 = (x−2)(y+1); denominator xy−2x+y−2 = (x+1)(y−2). Separating variables: ∫(y−2)/(y+1)dy = ∫(x−2)/(x+1)dx. Each integrand is 1 − 3/(·+1), giving y − 3ln|y+1| = x − 3ln|x+1| + c. Rearranging: x − y + 3ln|(y+1)/(x+1)| = c.
✓Final answerThe general solution is x − y + 3log|(y+1)/(x+1)| = c. The correct option is (C).
ANSWER: C
- TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQQ.The general solution of the differential equation (xsinxy)dy=(ysinxy−x)dx is (A) sin−1(xy)=2x+c (B) sin(yx)=2x2+c (C) sin(xy)=log∣x∣+c (D) cos(xy)=log∣x∣+c
›Reveal solutionSolution
Homogeneous DE; put y=vx, separate variables, and integrate to get cos(xy)=log∣x∣+c.
Write in standard form.
(xsinxy)dy=(ysinxy−x)dx⇒dxdy=xsinxyysinxy−x.
This is homogeneous (degree 0 in x,y).
Substitute y=vx, so dxdy=v+xdxdv and xy=v:
v+xdxdv=xsinvvxsinv−x=v−sinv1.
Separate variables.
xdxdv=−sinv1⇒sinvdv=−xdx.
Integrate.
−cosv=−log∣x∣+C⇒cosv=log∣x∣+c.
Back-substitute v=xy.
cos(xy)=log∣x∣+c.
✓Final answercos(xy)=log∣x∣+c — option (D).
- TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQQ.The general solution of the differential equation dxdy=2y3−4xy+y2y2+1 is (A) 4xy2+2x=y4+y2+c (B) 2xy2+x=y4−y2+c (C) 4xy2−2x=y4+y2+c (D) 4xy2+2x=y4−y2+c
›Reveal solutionSolution
Treat x as a function of y; the equation becomes linear in x, giving 4xy2+2x=y4+y2+c — option (A).
Step-by-step solution
Given dxdy=2y3−4xy+y2y2+1, invert to make x the dependent variable:
dydx=2y2+12y3−4xy+y.
Separate the x-term:
dydx+2y2+14yx=2y2+12y3+y=2y2+1y(2y2+1)=y.
This is linear in x. Integrating factor:
μ=e∫2y2+14ydy=elog(2y2+1)=2y2+1.
Then
dyd[x(2y2+1)]=y(2y2+1)=2y3+y.
Integrate:
x(2y2+1)=2y4+2y2+C.
Multiply through by 2:
4xy2+2x=y4+y2+c.
✓Final answer4xy2+2x=y4+y2+c — option (A).
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