Question

xy'=3y-6x^2
linear equation differential equation​

Answer 1

Working Rule :-

Let us consider a linear differential equation

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: \: where \: p \: and \: q \in \: f(x)$

Step :- 1 :- Find Integrating Factor, I. F.

$\rm :\longmapsto\:I. F. = {e}^{ \: \int p \: dx}$

Step :- 2, Solution of differential equation is

$\rm :\longmapsto\:y \times I. F. = \int \: (q \times I. F.) dx$

Formula Used :-

$\red{ \boxed{ \displaystyle \sf \: \int\dfrac{1}{x}dx = log(x) + c}}$

$\red{ \boxed{ \displaystyle \sf \: \int \: {x}^{n} dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}}$

$\red{ \boxed{ \sf \: {e}^{logx} = x}}$

Let's solve the problem now!!!

Given differential equation is

$\rm :\longmapsto\:x\dfrac{dy}{dx} = 3y - 6 {x}^{2}$

can be rewritten as

$\rm :\longmapsto\:x\dfrac{dy}{dx} - 3y = \: - \: 6 {x}^{2}$

On dividing by 'x' both sides, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} - \dfrac{3}{x} \: y = \: - \: 6 {x}$

So,

On comparing with

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: \:$

we get

$\red{\rm :\longmapsto\:p = - \: \dfrac{3}{x}} \\\red{\rm :\longmapsto\:q = - \: 6x}$

Integrating Factor

$\rm :\longmapsto\:I. F. = {e}^{ \int \: pdx}$

$\rm :\longmapsto\:I. F. = {e}^{ \displaystyle \int \: - \: \dfrac{3}{x} dx}$

$\rm :\longmapsto\:I. F. = {e}^{ - 3 \: logx}$

$\rm :\longmapsto\:I. F. = {e}^{\: log {x}^{ - 3} }$

$\rm :\longmapsto\:I. F. = {x}^{ - 3}$

Thus,

Solution of differential equation is

$\rm :\longmapsto\:y \times I. F. = \int \: (q \times I. F.) dx$

$\rm :\longmapsto\:y \times {x}^{ - 3} = \int \: ( - 6x \times {x}^{ - 3} ) dx$

$\rm :\longmapsto\:y \times {x}^{ - 3} = - 6 \: \int \: {x}^{ - 2} dx$

$\rm :\longmapsto\:y \times {x}^{ - 3} = - 6 \: \dfrac{ {x}^{ - 2 + 1} }{ - 2 + 1} + c$

$\rm :\longmapsto\:y \times {x}^{ - 3} = - 6 \: \dfrac{ {x}^{ - 1} }{ - 1} + c$

$\rm :\longmapsto\:y \times {x}^{ - 3} = \dfrac{6 }{x} + c$

$\rm :\longmapsto\:y = \dfrac{6 }{ {x}^{2} } + {c}{ {x}^{3}}$