Given: y2=cex−x−1Find the orthogonal family of curves and write 2 curves from both families that pass in (2,0).

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Makenzie Hart · Accepted Answer

Step 1To find the orthogonal family of curves, you need to first solve for c in terms of x and y.&amp;nbsp;y2=cex−x−12y∼y′=cex−1From the given family of curves,&amp;nbsp;c=1+x+y2exSo,&amp;nbsp;y′=x+y22ySlope for the orthogonal trajectories will be&amp;nbsp;∼y′=−2yx+y2Or,&amp;nbsp;∼2y∼&amp;nbsp;dx&amp;nbsp;+(x+y2)∼&amp;nbsp;dy&amp;nbsp;=0, which is of the form&amp;nbsp;M∼&amp;nbsp;dx&amp;nbsp;+N∼&amp;nbsp;dy&amp;nbsp;=0.Step 2To solve the above differential equation, note that it is not exact&amp;nbsp;(My≠Nx)&amp;nbsp;so first find an integrating factor that makes it exact, which in this case is&amp;nbsp;1y.Multiplying by the integrating factor,2y∼&amp;nbsp;dx&amp;nbsp;+(xy+y32)&amp;nbsp;dy&amp;nbsp;=0As&amp;nbsp;My=Nx, it is now exact and integrating, we get the solution2xy+25y52=CTo get two curves from both families that pass through the point (2,0), plug in&amp;nbsp;x=2,y=0&amp;nbsp;in the equation of curves and find values of c and C.

Given: \(\displaystyle{y}^{{2}}={c}{e}^{{x}}-{x}-{1}\) Find the orthogonal family of curves

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