The general solution of the differential equa | vedclass

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(B) Given the differential equation: $(2x - 10y^3) dy + y dx = 0$Rearranging the terms to form a linear differential equation in $x$:$y dx = (10y^3 -

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$x y^2 - 2y^5 = c$

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(B) Given the differential equation: $(2x - 10y^3) dy + y dx = 0$Rearranging the terms to form a linear differential equation in $x$:$y dx = (10y^3 - 2x) dy$$\frac{dx}{dy} = 10y^2 - \frac{2x}{y}$$\frac{dx}{dy} + \frac{2}{y} x = 10y^2$This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = \frac{2}{y}$ and $Q = 10y^2$.The integrating factor $(IF)$ is given by:$IF = e^{\int P dy} = e^{\int \frac{2}{y} dy} = e^{2 \ln y} = y^2$The general solution is $x \cdot (IF) = \int Q \cdot (IF) dy + C$:$x \cdot y^2 = \int (10y^2) \cdot y^2 dy + C$$x y^2 = \int 10y^4 dy + C$$x y^2 = 10 \cdot \frac{y^5}{5} + C$$x y^2 = 2y^5 + C$Thus, the general solution is $x y^2 - 2y^5 = C$.

The general solution of the differential equation $(2x - 10y^3) dy + y dx = 0, y \neq 0$ is

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