The real value of m for which the substitutio | vedclass

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(C) Given the differential equation: $2x^4y \frac{dy}{dx} + y^4 = 4x^6$. Substitute $y = u^m$,then $\frac{dy}{dx} = m u^{m-1} \frac{du}{dx}$. Substitu

Vedclass · Accepted Answer

$m = 3/2$

Vedclass · Answer

(C) Given the differential equation: $2x^4y \frac{dy}{dx} + y^4 = 4x^6$. Substitute $y = u^m$,then $\frac{dy}{dx} = m u^{m-1} \frac{du}{dx}$. Substituting these into the equation: $2x^4(u^m)(m u^{m-1} \frac{du}{dx}) + (u^m)^4 = 4x^6$. This simplifies to: $2m x^4 u^{2m-1} \frac{du}{dx} + u^{4m} = 4x^6$. For the equation to be homogeneous,the powers of $x$ and $u$ must be proportional in each term. Comparing the powers of $x$ and $u$ in the terms $x^4 u^{2m-1} \frac{du}{dx}$,$u^{4m}$,and $x^6$: From $u^{4m}$ and $x^6$,we require $4m = 6$,which gives $m = 3/2$. Checking the first term $x^4 u^{2m-1}$,we require $2m-1 = 2$,which also gives $m = 3/2$. Thus,the value of $m$ is $3/2$.

The real value of $m$ for which the substitution $y = u^m$ will transform the differential equation $2x^4y \frac{dy}{dx} + y^4 = 4x^6$ into a homogeneous equation is:

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