Q.Find the solution of dxdy=2y−x.
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.
Concept: Separation of Variables — rewrite the equation so all y terms are on one side and all x terms on the other, then integrate.
We have
dxdy=2y−x=2y⋅2−x.
Separate variables:
2−ydy=2−xdx.
Integrate both sides:
∫2−ydy=∫2−xdx.
Recall ∫a−udu=−lnaa−u+C. So
−ln22−y=−ln22−x+C.
Multiply through by −ln2:
2−y=2−x+C1(where C1=−Cln2).
Take reciprocal or solve for y:
2y=2−x+C11⇒y=log2(2−x+C1).
The solution is y=log2(2−x+C1).
Separable: 2−ydy=2−xdx. Integrating gives 2−x−2−y=C (equivalently 2−y−2−x=C).
Write 2y−x=2y2−x, so the equation separates:
2ydy=2−xdx⟹2−ydy=2−xdx.
Integrate both sides using ∫2−udu=−log22−u:
−log22−y=−log22−x+C.
Multiply through by −log2 and absorb constants:
2−y=2−x+C1⟹2−x−2−y=C.
Verification: differentiating 2−x−2−y=C gives −2−xlog2+2−ylog2dxdy=0, i.e. dxdy=2−y2−x=2y−x, as required.
The general solution is 2−x−2−y=C (equivalently 2−y−2−x=C), where C is an arbitrary constant.
Method: Variable-separable equations with an exponential right side
Use this whenever dxdy equals a product (or a power like ay−x) that can be split into a pure-x factor times a pure-y factor.
Steps
Step 1: Split the right side using index laws
A power such as ay−x separates as ay⋅a−x, turning the equation into
dxdy=(function of y)(function of x).
Step 2: Separate the variables
Move all y-terms (with dy) to one side and all x-terms (with dx) to the other:
function of ydy=(function of x)dx.
Step 3: Integrate both sides and add C
Recall ∫audu=lnaau. Integrate each side, keep exactly one arbitrary constant, and simplify to the general solution.
Common Mistakes
Mistake 1: Failing to split 2y−x into 2y⋅2−x
Why it's wrong: without splitting, the variables stay entangled and cannot be separated. Correct approach: use index laws first, then separate.
Mistake 2: Integrating 2−y as if it were e−y (dropping the ln2)
Why it's wrong: ∫audu=lnaau, so the ln2 factor is essential. Correct approach: keep the ln21 factors on both sides.
Mistake 3: Omitting the arbitrary constant
Why it's wrong: a general solution must carry one constant. Correct approach: add C after integrating.
- Higher Secondary (+2 Stage) Examination 2025Set ANNUAL1 markQ.Solve the differential equation dy=(1+y2)dx.
›Reveal solutionSolution
This is a variables-separable equation; separate and integrate, recognizing ∫1+y2dy=tan−1y.
Given dy=(1+y2)dx, separate the variables:
1+y2dy=dx.
Integrating both sides:
∫1+y2dy=∫dx ⇒ tan−1y=x+C.
Taking tangent of both sides gives the explicit solution:
y=tan(x+C).
✓Final answery=tan(x+C) (equivalently tan−1y=x+C).
- Higher Secondary (+2 Stage) Examination 2023Set ANNUAL1 markQ.Write the general solution of the differential equation dxdy=x5+x2−x2.
›Reveal solutionSolution
The equation dxdy=x5+x2−x2 is directly integrable term by term (variables already separated).
dy=(x5+x2−x2)dx
Integrating both sides:
y=∫x5dx+∫x2dx−2∫xdx
y=6x6+3x3−2ln∣x∣+C
✓Final answery=6x6+3x3−2ln∣x∣+C, where C is an arbitrary constant.
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