The solution of {y^5}x + y - xfrac{{dy}}{{dx} | vedclass

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(B) The given differential equation is ${y^5}x + y - x\frac{{dy}}{{dx}} = 0$.Rearranging the terms,we get ${y^5}x dx + y dx - x dy = 0$.Dividing the e

Vedclass · Accepted Answer

${x^5}/5 + (1/4){(x/y)^4} = C$

Vedclass · Answer

(B) The given differential equation is ${y^5}x + y - x\frac{{dy}}{{dx}} = 0$.Rearranging the terms,we get ${y^5}x dx + y dx - x dy = 0$.Dividing the entire equation by $x{y^5}$,we get $dx + \frac{{y dx - x dy}}{{x{y^5}}} = 0$.This can be rewritten as $dx + \frac{1}{{{x^4}}} \left( \frac{{y dx - x dy}}{{{y^2}}} \right) = 0$,which is $dx + {x^{ - 4}} d\left( {\frac{x}{y}} \right) = 0$.However,to match the standard form,let us multiply the original equation by $\frac{1}{{x^2{y^5}}}$: $\frac{1}{x} dx + \frac{{y dx - x dy}}{{x^2{y^5}}} = 0$.This is $\frac{1}{x} dx + \frac{1}{{{y^5}}} d\left( {\frac{x}{y}} \right) = 0$ is not quite right. Let us use the substitution $u = x/y$.Actually,dividing by $x^2 y^5$ gives $\frac{1}{x} dx + \frac{y dx - x dy}{x^2 y^5} = 0$,which is $d(\ln x) - \frac{1}{y^4} d(x/y)$ is not correct.Let us re-evaluate: ${y^5}x dx + y dx - x dy = 0$. Divide by $x^2 y^5$: $\frac{1}{x} dx + \frac{y dx - x dy}{x^2 y^5} = 0$. This is $\frac{1}{x} dx + \frac{1}{y^5} d(x/y) = 0$. This does not integrate easily.Let us divide by $x^2$: ${y^5} dx + \frac{y dx - x dy}{x^2} = 0$. This is ${y^5} dx + d(y/x) = 0$. Still not quite.Let us divide by $y^5$: $x dx + \frac{y dx - x dy}{y^5} = 0$. This is $x dx + \frac{1}{y^3} \frac{y dx - x dy}{y^2} = 0$. This is $x dx + (x/y)^{-3} d(x/y) = 0$ is wrong.Correct approach: ${y^5}x dx + y dx - x dy = 0$. Divide by $x^2 y^5$: $\frac{1}{x} dx + \frac{y dx - x dy}{x^2 y^5} = 0$. Let $u = x/y$,then $du = \frac{y dx - x dy}{y^2}$. So $\frac{1}{x} dx + \frac{1}{x^2 y^3} du = 0$. Actually,the solution provided in option $B$ is ${x^5}/5 + (1/4){(x/y)^4} = C$. Differentiating this: ${x^4} dx + (1/4) \cdot 4 {(x/y)^3} d(x/y) = 0 \Rightarrow {x^4} dx + {(x/y)^3} \frac{y dx - x dy}{y^2} = 0 \Rightarrow {x^4} dx + \frac{x^3}{y^5} (y dx - x dy) = 0$. Multiplying by $y^5/x^3$: $\frac{x^4 y^5}{x^3} dx + y dx - x dy = 0 \Rightarrow x{y^5} dx + y dx - x dy = 0$. This matches the original equation. Thus,option $B$ is correct.

The solution of ${y^5}x + y - x\frac{{dy}}{{dx}} = 0$ is

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