y32y5(x) =1-14x+Cy′5(x) =14(-14x+C)2= 14parenleftbigg1-14x+Cparenrightbigg2= 14y2After looking at the results of computingy′for each possible solution, we can see thatonlyy=y1solves the differential equation.Hence, the solution isy=1-7x+C.00810.0pointsThe family of solutions to the differentialequationy′=-8y2isy=18x+C.Find the solution that satisfies the initialconditiony(-5) = 8.1.y(x) =864(x+ 5) + 1correct2.y(x) =648(x+ 5) + 83.y(x) =164(x+ 5) + 8

feuge (ejf557) – Homework 4 – staron – (53940)54.y(x) =88(x+ 5) + 15.y(x) =18(x+ 5) + 8Explanation:This equation contains only one unknownconstant, so we will substitute the indicatedvalues ofxandyinto the equation to solveforCas follows:y(-5) =18(-5) +C=1C-40.This shows thaty(-5) = 8 if and only ifC=18+ 40. Thusy(x) =18x+18+ 40=18(x+ 5) +18.Hence,y(x) =864(x+ 5) + 1.00910.0pointsWhich of the following families of functions isthe solution to the differential equationy′= 6xy3?1.y=1C-6x22.y=1√C-6x2correct3.y=1C+ 6x24.y=1√C+ 6x25.y=6√C-6x2Explanation:This is a separable differential equation,but here we will solve it by the process ofelimination.y1(x) =1√C-6x2y′1(x) =6x(C-6x2)−12C-6x2=6x(C-6x2)32= 6xparenleftbigg1√C-6x2parenrightbigg3= 6xy3y2(x) =1C-6x2y′2(x) =12x(C-6x2)2= 12xparenleftbigg1C-6x2parenrightbigg2= 12xy2y3(x) =1√C+ 6x2y′3(x) =-6x(C+ 6x2)−12C+ 6x2=-6x(C+ 6x2)32=-6xparenleftbigg1√C+ 6x2parenrightbigg3=-6xy3y4(x) =1C+ 6x2y′4(x) =-12x(C+ 6x2)2=-12xparenleftbigg1C+ 6x2parenrightbigg2=-12xy2

feuge (ejf557) – Homework 4 – staron – (53940)6y5(x) =6√C-6x2y′5(x) =36x(C-6x2)−12C-6x2=36x(C-6x2)32= 36xparenleftbigg1√C-6x2parenrightbigg3= 36xy3After looking at the results of computingy′for each possible solution, we can see thatonlyy=y1solves the differential equation.Hence the solution isy=1√C-6x