Find a particular solution satisfying the given condition for differential equations. x dy\dx- y = log x, given that y = 0 when x = 1

Question

Find a particular solution satisfying the given condition for \(x \frac{dy}{dx}\) - y = log x, given that y = 0 when x = 1 differential equations.

x dy\dx- y = log x, given that y = 0 when x = 1

Answer 1

Given Differential Equation :

\(x \frac{dy}{dx}\) - y = log x

Formula :

i) \(\int \frac {1}{x}\) dx = log x

ii) alog b = log b^a

iii) a^{log_a b} = b

iv) \(\int\) u. v dx = u. \(\int\) v dx - \(\int\) \((\frac{du}{dx}. \int v, dx)\) dx

v) \(\int\) e^kx dx = \(\frac {e^{kx}}{k}\)

vi) \(\frac{d}{dx}\) 9kx) = k

vii) log 1 = 0

viii) 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,

Given differential equation is

Dividing above equation by x,

Equation (1) is of the form

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

Therefore, integrating factor is

General solution is

Put, log x =t => x=e^t

Therefore, (1/x) dx = dt

Let, u=t and v=e^-t

Substituting I in eq(2),

Multiplying above equation by x,

∴ y = - log x - 1 + cx

Therefore, general solution is

y = - log x - 1 + cx

For particular solution put y=0 and x=1 in above equation,