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A integrating factor of the differential equation dydx+yx=1x2\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x^2}, (x>0)(x > 0) is equal to

Solution

✅ Correct Option: 1

The given differential equation is:

dydx+yx=1x2\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x^2}

This is a first-order linear differential equation of the form dydx+P(x)⋅y=Q(x)\frac{dy}{dx} + P(x) \cdot y = Q(x)


Comparing with the standard form:

P(x)=1xP(x) = \frac{1}{x}

Q(x)=1x2Q(x) = \frac{1}{x^2}


The integrating factor is given by:

I.F.=e∫P(x)dxI.F. = e^{\int P(x)dx}

I.F.=e∫1xdxI.F. = e^{\int \frac{1}{x}dx}


Evaluating the integral:

∫1xdx=ln⁡∣x∣\int \frac{1}{x}dx = \ln|x|

Therefore:

I.F.=eln⁡∣x∣I.F. = e^{\ln|x|}

I.F.=∣x∣I.F. = |x|


Since x>0x > 0 is given, ∣x∣=x|x| = x

Therefore, the integrating factor is xx.

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