(a) Find the general solution of the differential equation x²(dy)/(dx) = x² + xy + y². OR (b) Find the particu

Question

x^2\dfrac{dy}{dx} = x^2 + xy + y^2

(OR)

(b) Find the particular solution of the differential equation

xy\dfrac{dy}{dx} = (x+2)(y+2)

, given that

y = -1

when

x = 1

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Part (a)Concept understanding — Homogeneous Differential Equation

Homogeneous Differential Equations

The idea: only the ratio $y/x$ matters

A function $f(x,y)$ is homogeneous of degree $n$ if scaling both variables by $t$ scales the function by $t^n$ : $f(tx,ty)=t^{n}f(x,y)$ . For example $f(x,y)=x^2+y^2$ gives $f(tx,ty)=t^2(x^2+y^2)$ — degree $2$ .

A first-order equation

\frac{dy}{dx}=f(x,y)

is called homogeneous when $f$ is homogeneous of degree $0$ , i.e. $f(tx,ty)=f(x,y)$ . Scaling then changes nothing, which means the slope depends only on the ratio $y/x$ , never on the absolute sizes. So a homogeneous equation can always be recast as

\frac{dy}{dx}=F\!\left(\frac{y}{x}\right).

Note

Quick test in the form $M(x,y)\,dx+N(x,y)\,dy=0$ : if every term of $M$ and $N$ has the same total degree (sum of the powers of $x$ and $y$ ), the equation is homogeneous. E.g. in $(x^2+y^2)\,dx-2xy\,dy=0$ both $M$ and $N$ are degree $2$ .

Why we care: it becomes separable

Recognising homogeneity buys you a guaranteed method. Substitute

y=vx \quad\Rightarrow\quad \frac{dy}{dx}=v+x\frac{dv}{dx}.

Putting this into $\dfrac{dy}{dx}=F(v)$ gives

v+x\frac{dv}{dx}=F(v)\quad\Rightarrow\quad x\frac{dv}{dx}=F(v)-v,

which separates:

\frac{dv}{F(v)-v}=\frac{dx}{x}.

Integrate both sides, then replace $v$ by $y/x$ to return to the original variables. …

Part (b)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:

\frac{dy}{dx} = f(x) \cdot 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:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). 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:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ 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 $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

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$\tan^{-1}(y/x)=\ln|x|+C$ .
$y-2\ln|y+2|=x+2\ln|x|-2$ .

Part (a)

Divide by $x^2$ : $\dfrac{dy}{dx}=1+\dfrac{y}{x}+\dfrac{y^2}{x^2}$ , homogeneous. Let $y=vx$ , $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ :

v+x\frac{dv}{dx}=1+v+v^2\ \Rightarrow\ x\frac{dv}{dx}=1+v^2.

Separate and integrate:

\int\frac{dv}{1+v^2}=\int\frac{dx}{x}\ \Rightarrow\ \tan^{-1}v=\ln|x|+C.

Restore $v=\dfrac{y}{x}$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 49 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$
›Reveal solutionSolution
This is a separable first-order ODE. Separate variables, integrate, and simplify to get $\sqrt{y} - \sqrt{x} = C$ , which is option (C).

The key idea: whenever you see $\frac{dy}{dx}$ expressed as a ratio of functions of $y$ and $x$ alone, you can "separate" them — move all $y$ terms to one side and all $x$ terms to the other — then integrate each side independently. That's exactly what we have here: $\frac{dy}{dx} = \frac{\sqrt{y}}{\sqrt{x}}$ .

Notice that $\sqrt{y}$ depends only on $y$ , and $\sqrt{x}$ only on $x$ . So the equation is already in separable form — we just need to rearrange it properly.

Separate the variables. Multiply both sides by $dx$ and divide by $\sqrt{y}$ :

$\frac{dy}{\sqrt{y}} = \frac{dx}{\sqrt{x}}$
This is valid as long as $x > 0$ and $y > 0$ (so the square roots are defined and non-zero).

Integrate both sides. Each side is a standard power integral:

$\int y^{-1/2}\, dy = \int x^{-1/2}\, dx$ $\frac{y^{1/2}}{1/2} = \frac{x^{1/2}}{1/2} + C_1$
which simplifies to:
$2\sqrt{y} = 2\sqrt{x} + C_1$

Simplify the constant. Divide through by 2:

$\sqrt{y} = \sqrt{x} + \frac{C_1}{2}$
Let $C = \frac{C_1}{2}$ (just renaming the arbitrary constant). Then:
$\sqrt{y} - \sqrt{x} = C$
Watch out

A common mistake is to forget the constant of integration or to combine the two integration constants incorrectly. When you integrate both sides, you get a constant on each side — but they can be merged into a single constant. Also, don't lose the factor of 2 from the power rule: $\int u^{-1/2}\, du = 2u^{1/2}$ , not $u^{1/2}$ . …
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CBSE 2026Set 65/3/11 markMCQ
$\frac{dy}{dx} = F(x, y)$
(i) $F(x, y) = 3x + 2y$
(ii) $F(x, y) = \sin \frac{y}{x} + \log y - \log x$
(iii) $F(x, y) = e^{y/x} + 1$
(iv) $F(x, y) = \sqrt{x^2 + y^2} - y$ (A)
(i) and
(ii) (B) (i),
(ii) and
(iii) (C) (ii),
(iii) and
(iv) (D)
(ii) and (iii)
›Reveal solutionSolution
A differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous exactly when $F$ is homogeneous of degree zero — meaning $F(tx, ty) = F(x, y)$ for all $t > 0$ , which is equivalent to $F$ being expressible purely as a function of $y/x$ . Only options (ii) and (iii) satisfy this, so the correct choice is (D).

The idea is simple: a homogeneous differential equation is one where the right-hand side $F(x, y)$ doesn't change if you scale both $x$ and $y$ by the same factor. Why does that matter? Because if $F$ has that property, you can substitute $y = vx$ and turn the equation into one in $v$ and $x$ alone — a separable equation you can actually solve. The test is clean: check whether $F(tx, ty) = F(x, y)$ for any $t > 0$ .

Let's go through each option.

Option (i): $F(x, y) = 3x + 2y$ Replace $x$ with $tx$ and $y$ with $ty$ :

$F(tx, ty) = 3(tx) + 2(ty) = t(3x + 2y) = t \cdot F(x, y)$
This is $t$ times the original, not equal to it — unless $t = 1$ . So $F$ is homogeneous of degree 1, not degree 0. It also cannot be written as a function of $y/x$ alone (try it: $3x + 2y = x(3 + 2(y/x))$ still has an $x$ factor outside). So this is not homogeneous for the purpose of $\frac{dy}{dx} = F(x, y)$ .

Option (ii): $F(x, y) = \sin\frac{y}{x} + \log y - \log x$ First simplify the log terms: $\log y - \log x = \log\frac{y}{x}$ . So

$F(x, y) = \sin\frac{y}{x} + \log\frac{y}{x}$
This is already written purely in terms of $y/x$ . That's a dead giveaway — it's homogeneous of degree 0. Check formally:
$F(tx, ty) = \sin\frac{ty}{tx} + \log\frac{ty}{tx} = \sin\frac{y}{x} + \log\frac{y}{x} = F(x, y)$
The $t$ cancels completely. So this is homogeneous.

Option (iii): $F(x, y) = e^{y/x} + 1$ Again, this is already a function of $y/x$ alone. …
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CBSE 2026Set 65/3/11 markMCQ
$\frac{dy}{dx} = e^{x+y}$
›Reveal solutionSolution
Separate variables in $\frac{dy}{dx} = e^{x+y}$ to find the general solution $e^x + e^{-y} = C$ ; the particular solution $e^x + e^{-y} = -2$ is impossible because the left side is always positive while the right is negative.

The differential equation $\frac{dy}{dx} = e^{x+y}$ is separable. The key insight is to rewrite the exponential sum in the exponent as a product: $e^{x+y} = e^x \cdot e^y$ . This lets us collect all $x$ -terms with $dx$ and all $y$ -terms with $dy$ .

Solving the differential equation

Separate the variables.

$\frac{dy}{dx} = e^x \cdot e^y$
Rearranging:
$\frac{dy}{e^y} = e^x \, dx$
or equivalently,
$e^{-y} \, dy = e^x \, dx$

Integrate both sides.

$\int e^{-y} \, dy = \int e^x \, dx$
The left side gives $-e^{-y}$ and the right gives $e^x$ :
$-e^{-y} = e^x + C_1$
where $C_1$ is an arbitrary constant.

Rearrange to standard form. Multiply through by $-1$ :

$e^{-y} = -e^x - C_1$
or equivalently,
$e^x + e^{-y} = -C_1$
Renaming $-C_1$ as $C$ (still an arbitrary constant):
$e^x + e^{-y} = C$
This is the general solution, so Reason (R) is true.

Checking the particular solution

Now examine the proposed particular solution $e^x + e^{-y} = -2$ .

For this to be valid, we need $C = -2$ in the general solution. But notice:

$e^x > 0$ for all real $x$ …
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CBSE 2026Set CX1 markMCQ
$\dfrac{dy}{dx}=e^{x+y}$
(a) $e^{-y}=e^{x}+c$
(b) $e^{x}+e^{-y}=c$
(c) $e^{-x}-e^{-y}=c$
(d) $e^{-x}+e^{-y}=c$
›Reveal solutionSolution

The equation is variable-separable; separating and integrating gives $e^{x}+e^{-y}=c$ — option (b).

Why separate? Since $e^{x+y}=e^{x}\,e^{y}$ , the right side factors into an $x$ -part and a $y$ -part, so the variables separate cleanly.

$\dfrac{dy}{dx}=e^{x}\,e^{y}\;\Rightarrow\;e^{-y}\,dy=e^{x}\,dx$

…

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CBSE 2026Set A1 markMCQ
$\frac{dy}{dx} = e^{x+y}$
(a) $e^x + e^{-y} = k$
(b) $e^x + e^y = k$
(c) $e^{-x} + e^y = k$
(d) $e^{-x} + e^{-y} = k$
›Reveal solutionSolution

Separate variables in $\tfrac{dy}{dx}=e^{x+y}$ : $\int e^{-y}dy=\int e^x dx\Rightarrow e^x+e^{-y}=k$ .

Write $\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$ and separate:

$e^{-y}\,dy=e^x\,dx$ .

…

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CBSE 2026Set A1 markMCQ
$x\frac{dy}{dx} = \cot y$
(a) $x\cos y = k$
(b) $x\tan y = k$
(c) $x\sec y = k$
(d) $x\sin y = k$
›Reveal solutionSolution

Separate variables: $\int\tan y\,dy=\int\tfrac{dx}{x}\Rightarrow \log|\sec y|=\log|x|+c\Rightarrow x\cos y=k$ .

From $x\dfrac{dy}{dx}=\cot y$ , separate:

$\dfrac{dy}{\cot y}=\dfrac{dx}{x}\Rightarrow \tan y\,dy=\dfrac{dx}{x}$ .

…

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CBSE 2026Set ANNUAL1 mark
$\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$
›Reveal solutionSolution

Separate variables and integrate both sides using the standard integral of $1/(1+t^2)$ .

$\dfrac{dy}{1+y^2} = \dfrac{dx}{1+x^2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x+y}$
(a) $e^x + e^{-y} = c$
(b) $e^x + e^y = c$
(c) $e^{-x} + e^y = c$
(d) $e^{-x} + e^{-y} = c$
›Reveal solutionSolution

Separate variables using $e^{x+y}=e^x\cdot e^y$ and integrate both sides.

$\dfrac{dy}{dx}=e^{x+y}=e^x\cdot e^y$

Separating variables: $e^{-y}\,dy=e^x\,dx$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Find the general solution of the differential equation \frac{dy}{dx} = (1 + x^2)(1 + y^2).$
›Reveal solutionSolution
$\tan^{-1}y=x+\dfrac{x^3}{3}+c$ .

Concept. A separable differential equation $\dfrac{dy}{dx}=g(x)h(y)$ is solved by collecting all $y$ -terms on one side and all $x$ -terms on the other, then integrating both sides.

Steps.

$\dfrac{dy}{dx}=(1+x^2)(1+y^2)$ .

Separate: $\dfrac{dy}{1+y^2}=(1+x^2)\,dx$ . …
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CBSE 2026Set ANNUAL1 markMCQ
$(x^{2}+1)\dfrac{dy}{dx}=1,\ y(1)=\dfrac{\pi}{2}$
(a) $y=\tan^{-1}x+\dfrac{\pi}{3}$
(b) $y=\tan^{-1}x$
(c) $y=\tan^{-1}x+\dfrac{\pi}{6}$
(d) $y=\tan^{-1}x+\dfrac{\pi}{4}$
›Reveal solutionSolution

Integrating gives $y=\tan^{-1}x+C$ ; the condition fixes $C=\tfrac{\pi}{4}$ .

Step 1: $(x^{2}+1)\dfrac{dy}{dx}=1\Rightarrow dy=\dfrac{dx}{x^{2}+1}$ .

Step 2: Integrating, $y=\tan^{-1}x+C$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{y\,dx - x\,dy}{y} = 0$
(a) $y = cx$
(b) $x = cy^2$
(c) $xy = c$
(d) $y = cx^2$
›Reveal solutionSolution

Separate variables and integrate.

$\dfrac{y\,dx-x\,dy}{y}=0 \implies y\,dx = x\,dy \implies \dfrac{dx}{x}=\dfrac{dy}{y}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$\dfrac{dy}{dx} = e^{x+y}$
(a) $e^x + e^y = C$
(b) $e^{-x} + e^y = C$
(c) $e^x + e^{-y} = C$
(d) $e^{-x} + e^{-y} = C$
›Reveal solutionSolution

Separate variables ( $e^{-y}dy = e^x dx$ ) and integrate both sides.

$\dfrac{dy}{dx} = e^{x+y} = e^x\cdot e^y$

Separating variables:

$e^{-y}\,dy = e^x\,dx$

Integrating both sides: …

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Q.(a) Find the general solution of the differential equation x2dydx=x2+xy+y2x^2\dfrac{dy}{dx} = x^2 + xy + y^2x2dxdy​=x2+xy+y2.

Part (a)

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Q.(a) Find the general solution of the differential equation $x^2\dfrac{dy}{dx} = x^2 + xy + y^2$ .

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