91影视

When the equation is of the form,$\frac{dy}{dx}=G\left(\frac{y}{x}\right)$, then the substitution$v=\frac{y}{x}$transforms the given equation into a separable equation in the variables v and x.

One has to solve the equation given in problem 7, i.e.,

$\left(y^3-胃y^2\right)d胃+2{胃}^{2}ydy=0\text{鈥�}\mathrm{........}\mathbf{\left(}1\mathbf{\right)}\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌�}$

Rewriting the equation (1) as,

$\begin{array}{l}\frac{dy}{d胃}=-\frac{\left(y^2-胃y\right)}{2{胃}^2}\\ \frac{dy}{d胃}=-\frac{1}{2}\left(\frac{y^2}{{胃}^2}-\frac{y}{胃}\right)\text{鈥�}\mathrm{......}\mathbf{\left(}\mathbf{2}\mathbf{\right)}\end{array}$

Substitute$t=\frac{y}{胃}$in equation (2),

So,$dt=\frac{胃dy-yd胃}{{胃}^2}$

${胃}^{2}dt=胃dy-yd胃$

Dividing both sides by $d胃$,

$\begin{array}{c}{胃}^2\frac{dt}{d胃}=胃\frac{dy}{d胃}-y\\ {胃}^2\frac{dt}{d胃}+y=胃\frac{dy}{d胃}\\ \frac{dy}{d胃}=胃\frac{dt}{d胃}+\frac{y}{胃}\\ \frac{dy}{d胃}=胃\frac{dt}{d胃}+t\end{array}$

Therefore, equation (2) becomes,

$\begin{array}{c}胃\frac{dt}{d胃}+t=-\frac{1}{2}\left(t^2-t\right)\\ 2胃\frac{dt}{d胃}+2t=-t^2+t\\ 2胃\frac{dt}{d胃}=-t^2+t-2t\\ 2胃\frac{dt}{d胃}=-t^2-t\end{array}$

$\frac{dt}{d胃}=\frac{-t\left(t+1\right)}{2胃}\text{鈥夆赌夆赌夆赌夆赌夆赌夆赌夆�夆��}\mathrm{......}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

Separate the variables in equation (3),

$\frac{-2}{t\left(t+1\right)}dt=\frac{1}{胃}d胃$

Integrating both sides,

$鈭�\frac{-2}{t\left(t+1\right)}dt=鈭�\frac{1}{胃}d胃鈰�鈰�鈰�鈰�鈰�鈰�\left(4\right)$

The left hand side of the equation (4) can be solved by using partial fraction,

Therefore,

$\frac{-2}{t\left(t+1\right)}=\frac{A}{t}+\frac{B}{t+1}$

Multiply both sides by $t\left(t+1\right)$,

$\begin{array}{c}-2=A\left(t+1\right)+Bt\text{鈥�}鈰�鈰�鈰�鈰�鈰�鈰�\left(5\right)\\ A=-2\end{array}$

Put t = -1 in equation (5),

$\begin{array}{c}-2=-B\\ B=2\end{array}$

Therefore, equation (4) becomes,

$鈭�\frac{-2}{t}dt+鈭�\frac{2}{t+1}dt=鈭�\frac{1}{胃}d胃$

Put t = 0 in equation (5),

$-2\ln \left(t\right)+2\ln \left(t+1\right)=\ln \text{鈥�}胃+\ln \text{鈥�}C$

Where, C is the constant of integration.

$\begin{array}{c}\ln {\left(t+1\right)}^2-\ln \left(t^2\right)=\ln \text{鈥�}+\ln \text{鈥�}C\\ \ln \left(\frac{{\left(t+1\right)}^2}{t^2}\right)=\ln \left(胃C\right)\\ \frac{{\left(t+1\right)}^2}{t^2}=胃C鈰�鈰�鈰�鈰�鈰�鈰�\left(6\right)\end{array}$

Put$t=\frac{y}{胃}$in equation (6),

${\left(\frac{y+胃}{y}\right)}^2=胃C$

Hence, the solution of the given equation $\mathbf{(}{\mathit{y}}^{\mathbf{3}}\mathbf{-}\mathit{胃}{\mathit{y}}^{\mathbf{2}}\mathbf{)}\mathit{d}\mathit{胃}\mathbf{+}\mathbf{2}{\mathit{胃}}^{\mathbf{2}}\mathit{y}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$is ${\left(\frac{y+胃}{y}\right)}^{\mathbf{2}}\mathbf{=}\mathit{胃}\mathit{C}$.