The solution of the differential equation x f | vedclass

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(B) Step $1$: Divide the equation by $xy^6$ to get $\frac{1}{y^6} \frac{dy}{dx} + \frac{1}{xy^5} = x^2$.Step $2$: Let $v = y^{-5}$, then $\frac{dv}{dx

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$y^{-5} = \frac{5}{2}x^3 + cx^5$

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(B) Step $1$: Divide the equation by $xy^6$ to get $\frac{1}{y^6} \frac{dy}{dx} + \frac{1}{xy^5} = x^2$.Step $2$: Let $v = y^{-5}$, then $\frac{dv}{dx} = -5y^{-6} \frac{dy}{dx}$, which implies $\frac{1}{y^6} \frac{dy}{dx} = -\frac{1}{5} \frac{dv}{dx}$.Step $3$: Substitute into the equation: $-\frac{1}{5} \frac{dv}{dx} + \frac{v}{x} = x^2$, which simplifies to $\frac{dv}{dx} - \frac{5}{x}v = -5x^2$.Step $4$: The integrating factor is $IF = e^{\int -\frac{5}{x} dx} = e^{-5 \ln x} = x^{-5}$.Step $5$: The solution is $v \cdot x^{-5} = \int (-5x^2) \cdot x^{-5} dx = \int -5x^{-3} dx = -5 \frac{x^{-2}}{-2} + c = \frac{5}{2}x^{-2} + c$.Step $6$: Multiplying by $x^5$, we get $v = \frac{5}{2}x^3 + cx^5$, so $y^{-5} = \frac{5}{2}x^3 + cx^5$.

The solution of the differential equation $x \frac{dy}{dx} + y = x^3y^6$ is:

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