cheatham (sc36975) – HW04 – um – (53890)23.y= 2e−5(x2−9)4.y= 2e5(x−3)25.y=e5(x2−9)+ 1Explanation:This equation contains only one unknownconstant, so we will substitute the indicatedvalues ofxandyinto the equation to solveforCas follows:y(3) =Ce5(3)2=Ce45.This shows thaty(3) = 2 if and only ifC= 2e−45. Thus,y(x) = 2e5x2e−45.Hence,y(x) = 2e5(x2−9).00410.0 pointsThe family of solutions to the differentialequationy′=−3y2isy=13x+C.Find the solution that satisfies the initialconditiony(−2) =−5.1.y(x) =53(x+ 2)−12.y(x) =515(x+ 2)−1correct3.y(x) =13(x+ 2)−54.y(x) =253(x+ 2)−55.y(x) =115(x+ 2)−5Explanation:This equation contains only one unknownconstant, so we will substitute the indicatedvalues ofxandyinto the equation to solveforCas follows:y(−2) =13(−2) +C=1C−6.This shows thaty(−2) =−5 if and only ifC=−15+ 6. Thusy(x) =13x−15+ 6=13(x+ 2)−15.Hence,y(x) =515(x