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Differential Equation| IIT JAM 2014 | Problem 4

Try this problem from IIT JAM 2014 exam. It requires knowledge of exact differential equation and partial derivative. We provide sequential hints.

Try this problem from IIT JAM 2014 exam. It requires knowledge of the exact differential equation and partial derivative.

Differential Equation | IIT JAM 2014 | Problem 4


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

  • $b=2,c=2a$
  • $b=4,c=2$
  • $b=2,c=4$
  • $b=2,a=2c$

Key Concepts


Differential Equation

Exact D/E

Partial Differentiation

Check the Answer


Answer: $b=4, c=2$

IIT JAM 2014 , Problem 4

Ordinary Differential Equations by Tenebaum and Pollard

Try with Hints


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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