Find the number of unique solutions of the following nonlinear ODEy′=103xy25,y(0)=−1

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entreblogsmc2j · Accepted Answer

Step 1Let&amp;nbsp;z=y15,hencey&amp;nbsp;″(x)=(z5)&amp;nbsp;″(x)=103xz(x)2=(5z4z′)′(x)=20z3(x)z′(x)2+5z(x)4z&amp;nbsp;″(x),which is equivalent to5z(x)2[z(x)2z&amp;nbsp;″(x)+4z(x)3z′(x)2−23x]=0,implying&amp;nbsp;z(x)=0&amp;nbsp;orz(x)2z&amp;nbsp;″(x)+4z(x)3z′(x)2−23x=0.I do not believe this latter equation can be solved analytically.Step 2Given the way you phrased the problem, and given how you went about mistakenly solving it, I assume you actually meant to write the differential equation asy&#039;(x)=103xy(x)25,instead. Then having only one boundary condition is justified in this situation. This can be solved with separation of variables. Hencey(x)=0&amp;nbsp;ory(x)-25y&#039;(x)=103x⇒53y35&#039;(x)=(53x2)&#039;⇒y35&#039;(x)=(x2)&#039;.By Riemann integrating, you obtainy(t)35-y(0)35=t2.ZSo in fact, you do obtain&amp;nbsp;y(t)=(t2-1)53,&amp;nbsp;and this is the only solution everywhere in&amp;nbsp;R

Find the number of unique solutions of the

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