Question

find the value of parameter a and b for which differential equation is
$(ay{x}^{2} + {y}^{3} )dx + (1 \div 3 {x}^{3} + bx {y}^{2} )dy$
is exact,and then solve it in the case of a and b.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm :\longmapsto\: ({ayx}^{2} + {y}^{3})dx + ( \dfrac{1}{3} {x}^{3} + {bxy}^{2})dy = 0$

On comparing with,

$\rm :\longmapsto\:Mdx \: + \: Ndy = 0$

we get

$\rm :\longmapsto\: M \: = \: {ayx}^{2} + {y}^{3}$

and

$\rm :\longmapsto\:N = \dfrac{1}{3} {x}^{3} + {bxy}^{2}$

Consider,

$\rm :\longmapsto\:\dfrac{\partial \: M}{\partial \: y} = {ax}^{2} + {3y}^{2}$

and

$\rm :\longmapsto\:\dfrac{\partial \: N}{\partial \: x} = {x}^{2} + {by}^{2}$

Now,

It is given that given Differential equation is exact Differential equation.

$\rm :\implies\:\dfrac{\partial \: M}{\partial \: y} = \dfrac{\partial \: N}{\partial \: x}$

$\rm :\longmapsto\: {ax}^{2} + {3y}^{2} = {x}^{2} + {by}^{2}$

So, on comparing we get,

$\rm :\longmapsto\:a = 1 \: \: \: \: and \: \: \: \: b = 3$

So,

Given Differential equation is rewritten as

$\rm :\longmapsto\: ({yx}^{2} + {y}^{3})dx + ( \dfrac{1}{3} {x}^{3} + {3xy}^{2})dy = 0$

Here,

$\rm :\longmapsto\: M \: = \: {yx}^{2} + {y}^{3}$

and

$\rm :\longmapsto\:N = \dfrac{1}{3} {x}^{3} + {3xy}^{2}$

We know,

Solution of exact Differential equation is given by

$\rm :\longmapsto\:\displaystyle\int _{y \: is \: constant} \: Mdx \: + \: \displaystyle\int _{term \: not \: containing \: x}Ndy = c$

$\rm :\longmapsto\:\displaystyle\int _{y \: is \: constant} \: ( {yx}^{2} + {y}^{3}) dx \: + \: \displaystyle\int _{term \: not \: containing \: x}0 \: dy = c$

$\rm :\longmapsto\:\dfrac{y {x}^{3} }{3} + {y}^{3}x = c$