Find a particular solution satisfying the given condition x dy\dx + y = x^3 , given that y =1 when x =2 differential equations.

Question

Find a particular solution satisfying the given condition for \(x \frac{dy}{dx}\) + y = x³ , given that y =1 when x =2 differential equations.

x dy\dx + y = x³ , given that y =1 when x =2

Answer 1

Given Differential Equation :

\(x \frac{dy}{dx}\) + y = x³ 

Formula :

i) \(\int\) \(\frac{1}{x}\) dx = log x

ii) a^{log_a b} =b

iii) \(\int\) xⁿ dx = \(\frac{x^{n+1}}{n+1}\)

iv) General solution :

For the differential equation in the form of

\(\frac{dy}{dx} \, + Py\, =Q\)

General solution is given by,

y. (I.F.) = \(\int\) Q. (I.F.) dx + c

Where, integrating factor,

I. F. = \(e^{\int p\, dx}\)

Given differential equation is

 \(x \frac{dy}{dx}\) + y = x³ 

Dividing above equation by x,

Equation (1) is of the form

\(\frac{dy}{dx} \, + Py\, =Q\)

where, P = \(\frac{1}{x}\) and Q = x²

Therefore, integrating factor is

General solution is

Dividing above equation by x,

Therefore general equation is

For particular solution put y=1 and x=2 in above equation,

Therefore, particular solution is