Question

Find the particular solution of
3xdy/dx- y = log x +1.x>0
​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The solution of the given Differential Equation is

$\boxed{y=-(\log x+4)+Cx^{1/3}}$

Step-by-step explanation:

Given Differential equation

$3x\frac{dy}{dx}-y=\log x+1$

$\implies \frac{dy}{dx}-\frac{y}{3x}=\frac{\log x+1}{3x}</p><p>Integrating factor</p><p>[tex]=e^{\int (-1/3x)dx}=x^{-1/3}$

Multiplying the differential equation by the integrating factor

$\frac{dy}{dx}\times x^{-1/3}-\frac{y}{3x}\times x^{-1/3}=\frac{\log x+1}{3x}\times x^{-1/3}$

$\implies \frac{d}{dx}(yx^{-1/3})=x^{-4/3}(\log x+1)/3$

$\implies \int d(yx^{-1/3})=\int x^{-4/3}(\log x+1)/3 dx$

$\implies 3yx^{-1/3}=(\log x+1)\int x^{-4/3}dx-\int (\frac{1}{x}\times \int x^{-4/3}dx)dx$

$\implies 3yx^{-1/3}=-3(\log x+1)x^{-1/3}-\int (\frac{1}{x}\times (-3x^{-1/3}))dx$

$\implies 3yx^{-1/3}=-3(\log x+1)x^{-1/3}+3\int x^{-4/3}dx$

$\implies 3yx^{-1/3}=-3(\log x+1)x^{-1/3}-9x^{-1/3}$

$\implies yx^{-1/3}=-(\log x+1)\int x^{-1/3}-3x^{-1/3}$

$\implies yx^{-1/3}=-x^{-1/3}(\log x+4)+C$

$\implies y=-(\log x+4)+Cx^{1/3}$

This is the general solution of the given differential equation.

For particular solution some condition must be given.

Hope this answer is helpful.

Know More:

Q: The solution of the differential equation dy/dx = (x + y)/x satisfying the condition y(1) = 1 is :

Click Here: https://brainly.in/question/5231429

Q: Find the particular solution of the differential equation : (x-siny)dy+(tany)dx=0 given that y=0 when x=0.

Click Here:  https://brainly.in/question/8157527