Q.Solve the following differential equation: x(x2−1)dxdy=1;y=0 when x=2
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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 …
Separable — integrate dy=x(x2−1)dx using partial fractions.
Partial fractions: x(x−1)(x+1)1=−x1+2(x−1)1+2(x+1)1.
Integrate:
y=−log∣x∣+21log∣x−1∣+21log∣x+1∣+C=21logx2x2−1+C.
Apply y=0 at x=2: 0=21log43+C⇒C=21log34.
So …
Separate, integrate by partial fractions, and use y(2)=0: y=21log(3x24(x2−1)).
1. Separate the variables
x(x2−1)dxdy=1⇒dy=x(x2−1)dx=x(x−1)(x+1)dx.
2. Partial fractions
Write x(x−1)(x+1)1=xA+x−1B+x+1C. Covering each factor and substituting its root:
A=(−1)(1)1=−1,B=(1)(2)1=21,C=(−1)(−2)1=21.
3. Integrate
y=∫(−x1+2(x−1)1+2(x+1)1)dx=−log∣x∣+21log∣x−1∣+21log∣x+1∣+C.
Combine the last two logs, 21log∣x−1∣+21log∣x+1∣=21log∣x2−1∣, so
y=21logx2x2−1+C.
4. Use the condition y=0 at x=2 …
Method: Separation with partial fractions over three linear factors
Use this when separation leaves x(x−1)(x+1)dx-type integrals — a product of distinct linear factors that partial fractions handle.
Steps
Step 1: Separate
Isolate dy so that dy=x(x2−1)dx=x(x−1)(x+1)dx.
Step 2: Split into simple fractions
Write x(x−1)(x+1)1=xA+x−1B+x+1C and solve by covering-up each factor. Integrate to a sum of logarithms. …
Common Mistakes
Mistake 1: Not factoring x2−1 before decomposing
Why it's wrong: partial fractions need the fully factored denominator x(x−1)(x+1); leaving x(x2−1) hides the linear factors. Correct approach: factor first, then split.
Mistake 2: Sign/coefficient errors in the partial fractions
Why it's wrong: the correct split is −x1+x−11/2+x+11/2; a wrong A,B,C yields the wrong log combination. Correct approach: use the cover-up rule at x=0,1,−1. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.The solution of the differential equation (2y−1)dy−(y−2)dx=0 is (A) 2y+3log∣y−2∣=2x+c (B) 2y+4log∣y−2∣=x+c (C) y+3log∣y−2∣=x+c (D) 2y+3log∣y−2∣=x+c (E) 3y+2log∣y−2∣=x+c
›Reveal solutionSolution
Separate variables and divide 2y−1 by y−2.
(2y−1)dy=(y−2)dx⇒dx=y−22y−1dy.
y−22y−1=y−22(y−2)+3=2+y−23. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.The general solution of the differential equation (x+y)2dxdy=1 is (A) y=21tan−1(x+y)+c (B) y=−(x+y)−1+c (C) y=31(x+y)3+c (D) y=sin−1(x+y)+c (E) y=tan−1(x+y)+c
›Reveal solutionSolution
Put v=x+y; the equation becomes 1+v2v2dv=dx, whose integral v−tan−1v=x+c simplifies to y=tan−1(x+y)+c.
Let v=x+y, so dxdv=1+dxdy, i.e. dxdy=dxdv−1. Substitute into (x+y)2dxdy=1:
v2(dxdv−1)=1 ⇒ v2dxdv=1+v2.
Separate variables: …
- KEAM 2025Set eng-2025-04254 marksMCQQ.The general solution of the differential equation (1+y)dx−(1−x)dy=0 is (A) x2+y2+x−y=C (B) x+y−xy=C (C) x−y+xy=C (D) x−y−xy=C (E) x2−y2+x+y=C
›Reveal solutionSolution
Separate variables, integrate, and simplify the resulting product to the given form.
From (1+y)dx=(1−x)dy,
1−xdx=1+ydy.
Integrating,
−loge∣1−x∣=loge∣1+y∣+c⇒loge(1−x)(1+y)=const, …
- KEAM 2024Set eng-2024-06064 marksMCQQ.The general solution of the differential equation dxdy=xy−2x−2y+4 is (A) (y−2)21=2(x−2)2+C (B) loge∣y−2∣=2(x−2)2+C (C) (y−2)2=2(x−2)2+C (D) loge∣y−2∣=C (E) loge∣y−2∣=(x−2)2+C
›Reveal solutionSolution
The equation is separable: (x−2)(y−2), giving loge∣y−2∣=2(x−2)2+C.
Factor the RHS: xy−2x−2y+4=x(y−2)−2(y−2)=(x−2)(y−2).
Separate variables: y−2dy=(x−2)dx. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.The solution of (ey+1)cosxdx+eysinxdy=0 is (A) (ey+1)sinx=C (B) (ey+1)=Csinx (C) ey=Csinx (D) (ey−1)sinx=C (E) (ey+1)cosx=C
›Reveal solutionSolution
Separate and integrate: log(ey+1)+logsinx=const, giving (ey+1)sinx=C.
(ey+1)cosxdx+eysinxdy=0⇒eysinxdy=−(ey+1)cosxdx.
Separate: ey+1eydy=−sinxcosxdx. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.The solution of cosydy=dx is (A) log∣secy−tany∣=x+C (B) x+secy+tany=C (C) secy+tany=x+C (D) log∣secx+tany∣=secy+x+C (E) log∣secy+tany∣=x+C
›Reveal solutionSolution
Separate and use ∫secydy=log∣secy+tany∣.
Rewrite cosydy=dx as secydy=dx. Integrating both sides: …
- KEAM 2024Set eng-2024-06054 marksMCQQ.The solution of (ycosy+siny)dy=(2xlogx+x)dx is (A) ysinx=x2logx+C (B) ysiny=xlogx+C (C) ysiny=x2logx+C (D) sinx=x2logx+C (E) ysinx=xlogx+C
›Reveal solutionSolution
Recognize d(ysiny) on the left and integrate the right to x2logx.
The left side is exact: dyd(ysiny)=siny+ycosy, so ∫(ycosy+siny)dy=ysiny. For the right: …
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.The general solution of the differential equation 4xy+12x+(2x2+3)y′=0 is (A) y+32x2+3=C (B) 2x2+3y−3=C (C) 2x2+3y+2=C (D) (y−3)(2x2+3)=C (E) (y+3)(2x2+3)=C
›Reveal solutionSolution
The general solution is (y+3)(2x2+3)=C.
Concept and Intuition
The equation is separable once grouped: the y-terms factor as y+3 and the x-terms give a logarithm whose argument is 2x2+3.
Step-by-Step Solution
- 4xy+12x+(2x2+3)y′=0⇒(2x2+3)y′=−4x(y+3).
- Separate: y+3dy=−2x2+34xdx.
- Integrate: log∣y+3∣=−log∣2x2+3∣+c (since ∫2x2+34xdx=log∣2x2+3∣). …
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