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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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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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