How to solve the differential equationy′=2y(xy−1)

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disolutoxz61 · Accepted Answer

Step 1I&#039;ll let z=y. Then z2=y, so dydx=2zdzdxYour DE turns into 2zz′=2z2(xz−1), z′=xz2−z, 1z+z′z2=xLet v=1z, then −v′=z′z2v−v′=x, so −(e−xv)′=e−xv−e−xv′=xe−xv=ex∫{−xe−x}dx=ex((x+1)e−x+C)=Cex+x+1y=z2=1v2=1(Cex+x+1)2

Ireland Vaughan · Accepted Answer

Step 1Well, we have the following DE:1)&amp;nbsp;y′(x)=ny(x)(xy(x)−k)We can rewrite the equation:2)&amp;nbsp;y′(x)+kny(x)=nxy(x)32Dividing both sides&amp;nbsp;−2y(x)323)&amp;nbsp;−y′(x)2y(x)32−kn2y(x)=−nx2Let&amp;nbsp;v(x)=1y(x),&amp;nbsp;which gives&amp;nbsp;v′(x)=−y′(x)2y(x)324)&amp;nbsp;v′(x)−knv(x)2=−nx2Now, let&amp;nbsp;μ(x)=exp⁡(∫−kn2dx)=exp⁡(−knx2). Multiply both sides by&amp;nbsp;μ(x), substitute&amp;nbsp;−knexp⁡(−knx2)2=ddx(exp⁡(−knx2))&amp;nbsp;and apply the reverse product rule:5)&amp;nbsp;ddx(exp⁡(−knx2)v(x))=−nxexp⁡(−knx2)2Integrate both sides with respect to x, evaluate the integrals and divide both sides by&amp;nbsp;μ(x). Now you can solve for y(x) by using&amp;nbsp;y(x)=1v(x)2

How to solve the differential equation \(\displaystyle{y}'={2}{y}{\left({x}\sqrt{{{y}}}-{1}\right)}\)

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