solve the given initial - value problem. the de is homogeneous.\n\\(\\frac{dy}{dx}=\\frac{y^{3}-x^{3}}{xy^{2}…

Explanation:

Step1: Rewrite the differential equation

Given $\frac{dy}{dx}=\frac{xy^{2}}{y^{3}-x^{3}}$, rewrite it as $\frac{dy}{dx}=\frac{\frac{y^{2}}{x^{2}}}{\frac{y^{3}}{x^{3}} - 1}$. Let $v = \frac{y}{x}$, then $y=vx$ and $\frac{dy}{dx}=v + x\frac{dv}{dx}$. Substituting these into the differential - equation, we get $v + x\frac{dv}{dx}=\frac{v^{2}}{v^{3}-1}$.

Step2: Separate the variables

Rearrange the equation: [ \begin{align*} x\frac{dv}{dx}&=\frac{v^{2}}{v^{3}-1}-v\ x\frac{dv}{dx}&=\frac{v^{2}-v(v^{3}-1)}{v^{3}-1}\ x\frac{dv}{dx}&=\frac{v^{2}-v^{4}+v}{v^{3}-1}\ \frac{v^{3}-1}{-v^{4}+v^{2}+v}dv&=\frac{dx}{x} \end{align*} ] Factor the denominator of the left - hand side: $-v^{4}+v^{2}+v=-v(v^{3}-v - 1)$. We can also rewrite the left - hand side as $\frac{v^{3}-1}{v(-v^{3}+v + 1)}dv=\frac{dx}{x}$. Another way is to use partial fraction decomposition on $\frac{v^{3}-1}{-v^{4}+v^{2}+v}$. First, factor out $v$ from the denominator: [ \begin{align*} \frac{v^{3}-1}{v(-v^{3}+v + 1)}dv&=\frac{dx}{x}\ \int\frac{v^{3}-1}{v(-v^{3}+v + 1)}dv&=\int\frac{dx}{x} \end{align*} ] A more straightforward approach is to rewrite the original homogeneous equation as $\frac{y^{3}-x^{3}}{xy^{2}}dy = dx$. Let $y = vx$, then $dy=v dx+x dv$. [ \begin{align*} \frac{(vx)^{3}-x^{3}}{x(vx)^{2}}(v dx+x dv)&=dx\ \frac{v^{3}x^{3}-x^{3}}{x\cdot v^{2}x^{2}}(v dx+x dv)&=dx\ \frac{x^{3}(v^{3}-1)}{v^{2}x^{3}}(v dx+x dv)&=dx\ \frac{v^{3}-1}{v^{2}}(v dx+x dv)&=dx\ \left(\frac{v^{4}-v}{v^{2}}\right)dx+\frac{v^{3}-1}{v^{2}}x dv&=dx\ \left(\frac{v^{4}-v}{v^{2}} - 1\right)dx+\frac{v^{3}-1}{v^{2}}x dv&=0\ \left(\frac{v^{4}-v - v^{2}}{v^{2}}\right)dx+\frac{v^{3}-1}{v^{2}}x dv&=0\ \frac{v^{4}-v^{2}-v}{v^{2}}dx=-\frac{v^{3}-1}{v^{2}}x dv\ \frac{v^{4}-v^{2}-v}{v^{3}-1}dv=-\frac{dx}{x} \end{align*} ] Integrating both sides: [ \begin{align*} \int\frac{v^{4}-v^{2}-v}{v^{3}-1}dv&=-\int\frac{dx}{x}\ \int\left(v+\frac{v}{v^{3}-1}\right)dv&=-\int\frac{dx}{x}\ \int vdv+\frac{1}{3}\int\frac{3v}{v^{3}-1}dv&=-\int\frac{dx}{x}\ \frac{v^{2}}{2}+\frac{1}{3}\ln|v^{3}-1|&=-\ln|x| + C \end{align*} ] Substitute back $v=\frac{y}{x}$: [ \frac{y^{2}}{2x^{2}}+\frac{1}{3}\ln\left|\frac{y^{3}}{x^{3}}-1\right|=-\ln|x| + C ]

Step3: Use the initial condition

Given $y(1) = 3$, substitute $x = 1$ and $y = 3$ into the above equation: [ \frac{3^{2}}{2\times1^{2}}+\frac{1}{3}\ln\left|\frac{3^{3}}{1^{3}}-1\right|=-\ln|1|+C ] [ \frac{9}{2}+\frac{1}{3}\ln(27 - 1)=C ] [ \frac{9}{2}+\frac{1}{3}\ln(26)=C ] The solution of the initial - value problem is $\frac{y^{2}}{2x^{2}}+\frac{1}{3}\ln\left|\frac{y^{3}}{x^{3}}-1\right|=-\ln|x|+\frac{9}{2}+\frac{1}{3}\ln(26)$.

Answer:

$\frac{y^{2}}{2x^{2}}+\frac{1}{3}\ln\left|\frac{y^{3}}{x^{3}}-1\right|=-\ln|x|+\frac{9}{2}+\frac{1}{3}\ln(26)$