Question

solve linear differential equation
(4x^2y-6)dx+x^3dy=0​

Answer 1

Answer:

sorry don't know soerrrry

Answer 2

Concept

Consider the differential equation

dy/dx+Py=Q, where P and Q are the functions of x only. Then the solution of such differential equation is given as,

y*IF=Integration[Q*IF] + C. where IF=exp(integration[Pdx]) and C is the integration constant.

Given

The given differential equation is (4x^2y-6)dx+x^3dy=0​.

Find

We have to calculate the solution of the given differential equation.

Solution

Since,

(4x^2y-6)dx+x^3dy=0​

Writing this in the the general form of linear differential equation i.e.

dy/dx + 4y/x=6/x^3

Therefore comparing this with dy/dx+Py=Q, we have

P=4/x and Q=6/x^3

Therefore the integrating factor will be,

$IF=e^{\int {4/x} \, dx } \\=e^{4lnx}\\=x^4$

Therefore the solution will be,

$yx^4=\int {\dfrac{6}{x^{3}}x^4} \, dx+C\\yx^4=\int {6x} \, dx+C\\yx^4=3x^2+C$

Hence the solution of the given linear differential equation is yx^4=3x^2+C.

#SPJ2