Q.For the differential equation xydxdy=(x+2)(y+2), find the solution curve passing through the point (1,−1).
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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 …
Concept: Separation of Variables
We rewrite the equation so that all y terms are with dy and all x terms with dx:
y+2ydy=xx+2dx
Integrate both sides. The left side: ∫y+2ydy=∫(1−y+22)dy=y−2log∣y+2∣+C1.
The right side: ∫xx+2dx=∫(1+x2)dx=x+2log∣x∣+C2.
Combining constants gives the general solution:
y−2log∣y+2∣=x+2log∣x∣+C …
The solution curve through (1,−1) is y=x−2+2log∣x(y+2)∣.
The equation xydxdy=(x+2)(y+2) is separable.
Separate the variables.
y+2ydy=xx+2dx.
Simplify each side. Using y+2y=1−y+22 and xx+2=1+x2:
(1−y+22)dy=(1+x2)dx.
Integrate both sides.
y−2log∣y+2∣=x+2log∣x∣+C.
Apply the point (1,−1). With x=1,y=−1: −1−2ln1=1+2ln1+C, so C=−2.
Hence y−2log∣y+2∣=x+2log∣x∣−2, i.e. …
Method: Separation via splitting each rational side
Use this when a curve's equation like xyy′=(x+2)(y+2) separates into y+2ydy and xx+2dx, each simplified by polynomial-style splitting.
Steps
Step 1: Separate
Group so that y+2ydy=xx+2dx.
Step 2: Rewrite each quotient as "1 minus/plus a simple fraction"
Use y+2y=1−y+22 and xx+2=1+x2, then integrate to linear-plus-log expressions.
Step 3: Apply the point to fix the constant …
Common Mistakes
Mistake 1: Integrating y+2y or xx+2 directly
Why it's wrong: these are improper-looking ratios; they must first be rewritten as 1−y+22 and 1+x2. Correct approach: split off the whole part before integrating.
Mistake 2: Wrong sign on the y+22 term …
- GUJCET 2024Set 131 markMCQQ.The general solution of the differential equation yxdy−ydx=0 is __________. (A) y=cx2 (B) x=cy2 (C) y=cx (D) xy=c
›Reveal solutionSolution
The equation reduces to ydy=xdx, whose solution is y=cx.
Steps. From yxdy−ydx=0 we get xdy−ydx=0⇒xdy=ydx, so
ydy=xdx. …
- GUJCET 2026Set x1 markMCQQ.The general solution of the differential equation dxdy=ex+y is ______ (A) ex+e−y=C (B) e−x+ey=C (C) ex+ey=C (D) e−x+e−y=C
›Reveal solutionSolution
Separate variables in dxdy=exey to get ex+e−y=C.
Write dxdy=ex+y=exey and separate:
e−ydy=exdx
Integrating both sides:
−e−y=ex+c1 …
- GUJCET 2025Set 031 markMCQQ.The general solution of the differential equation dxdy=ex−y is _____. (A) e−x−e−y=c (B) ex−ey=c (C) e−x−ey=c (D) ex−e−y=c
›Reveal solutionSolution
[!TLDR]
B increases linearly inside the wire up to r=a, then decreases as 1/r outside — graph (B).
Concept
Using Ampere's law for a long straight wire of radius a carrying uniformly distributed current I:
- Inside (r<a): B=2πa2μ0Ir — proportional to r.
- Outside (r>a): B=2πrμ0I — proportional to 1/r.
Solution …
- GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQQ.The general solution of the differential equation dxdy=ex−y is ____.(a) ey−ex=C(b) ex+e−y=C(c) ey+ex=C(d) e−x+ey=C
›Reveal solutionSolution
This equation separates directly once ex−y is split into ex⋅e−y.
dxdy=ex⋅e−y⇒eydy=exdx.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.The general solution of a differential equation yydx−xdy=0 is ______.(a) y=Cx2(b) y=Cx(c) x=Cy2(d) xy=C
›Reveal solutionSolution
Separate the variables directly.
ydx−xdy=0⇒xdx=ydy.
…
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.The general solution of the differential equation dxdy=1+x21+y2 is ___.(a) tan−1y=tan−1x+C(b) sin−1y=sin−1x+C(c) log∣y2+1∣=log∣1+x2∣+C(d) cos−1y=cos−1x+C
›Reveal solutionSolution
Separate variables: 1+y2dy=1+x2dx.
dxdy=1+x21+y2⇒∫1+y2dy=∫1+x2dx.
…
- GUJCET 2021Set 151 markMCQQ.The general solution of the differential equation dxdy=ex−y is (A) ex+ey=C (B) e−x+ey=C (C) e−x+e−y=C (D) ex−ey=C
›Reveal solutionSolution
The equation is variable-separable via e^(x-y) = e^x e^-y.
Concept. dxdy=ex−y=exe−y⇒eydy=exdx. …
- GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQQ.The solution of the differential equation 2xdxdy−y=0; y(1)=2 represents ______.(a) Parabola(b) Straight line(c) Circle(d) Ellipse
›Reveal solutionSolution
Separate variables, integrate, and apply y(1)=2 to identify the curve.
2xdxdy=y⇒ydy=2xdx. Integrating: logy=21logx+C⇒y=kx.
…
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