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Differential Equation| IIT JAM 2014 | Problem 4

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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