Try this problem from IIT JAM 2014 exam. It requires knowledge of the exact differential equation and partial derivative.
For $a,b,c \in \mathbb R$, If the differential equation $(ax^2+bxy+y^2)\mathrm d x+(2x^2+cxy+y^2)\mathrm d y=0$ is exact then
Differential Equation
Exact D/E
Partial Differentiation
Answer: $b=4, c=2$
IIT JAM 2014 , Problem 4
Ordinary Differential Equations by Tenebaum and Pollard
Any first order and linear differential equation can be expressed as
$M(x,y)\mathrm d x+N(x,y)\mathrm d y=0 \cdots\cdots\cdots (i)$ where $M$ and $N$ are functions of $x$ and $y$.
$(i)$ is called exact if and only if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Comparing the given equation with $(i)$ we get,
$M(x,y)=(ax^2+bxy+y^2)$
$N(x,y)=(2x^2+cxy+y^2)$
Now can you calculate $\frac{\partial M}{\partial y}\text{ and }\frac{\partial N}{\partial x}$ ?
$\frac{\partial M(x,y)}{\partial y}=2y+bx$ [differentiating $M(x,y)$ with respect to $y$ taking $x$ as constant.]
$\frac{\partial N(x,y)}{\partial x}=4x+cy$ [differentiating $N(x,y)$ with respect to $x$ taking $y$ as constant.]
Since the equation is exact ,then$ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $
i.e., $2y+bx=4x+cy$
Comparing the coefficients we get $b=4,c=2$ [ANS]
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