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The general solution of the differential equation xdydx+4y=x3,(x≠0)\frac{xdy}{dx} + 4y = x^3, (x \neq 0) is:

Solution

✅ Correct Option: 1

The differential equation is xdydx+4y=x3,(x≠0)\frac{xdy}{dx} + 4y = x^3, (x \neq 0)

Dividing the entire equation by xx:

dydx+4yx=x3x\frac{dy}{dx} + \frac{4y}{x} = \frac{x^3}{x}

dydx+4xy=x2\frac{dy}{dx} + \frac{4}{x}y = x^2

This gives P(x)=4xP(x) = \frac{4}{x} and Q(x)=x2Q(x) = x^2


The integrating factor is:

I.F.=e∫P(x)dxI.F. = e^{\int P(x)dx}

I.F.=e∫4xdxI.F. = e^{\int \frac{4}{x}dx}

I.F.=e4ln⁡xI.F. = e^{4\ln x}

I.F.=eln⁡x4I.F. = e^{\ln x^4}

I.F.=x4I.F. = x^4


The general solution is given by:

y×I.F.=∫Q(x)×I.F. dx+cy \times I.F. = \int Q(x) \times I.F. \, dx + c

y×x4=∫x2×x4 dx+cy \times x^4 = \int x^2 \times x^4 \, dx + c

y⋅x4=∫x6 dx+cy \cdot x^4 = \int x^6 \, dx + c

y⋅x4=x77+cy \cdot x^4 = \frac{x^7}{7} + c


Dividing both sides by x4x^4:

y=x77x4+cx4y = \frac{x^7}{7x^4} + \frac{c}{x^4}

y=x37+cx−4y = \frac{x^3}{7} + cx^{-4}

Therefore, the general solution is y=x37+cx−4y = \frac{x^3}{7} + cx^{-4} where cc is an arbitrary constant.

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