Question

#4 Solve the following: (1 point) Solve the differential equation 6y 2 +2 where y 6 when 0 (1 point) The differential equation can be written in differential form: M(x, y) dz +N(z, ) dy-0 where ,and N(x, y)--y5-3x The term M(, y) dz + N(x, y) dy becomes an exact differential if the left hand side above is divided by y4. Integrating that new equation, the solution of the differential equation is E C

Answer 1

For differential equation y' + p(x) y = Qx Integrating factor is I F = e6integral p(x) dx and the solution is, y I F = integral Q(x) I f dx here P(x) = 6/x + 2 I f = e^integral 6/x + 2 dx = e^6 ln (x + 2) = e^ln (x + 2)^6 = (x + 2)^6 therefore y(x) (x + 2)^6 = integral (x + 2)^6 (x + 2)^6 dx = integral (x + 2)^12 dx = (x + 2)^12/12 + c therefore y(x) = (x + 2)^7/13 + c/(x + 2)^6 therefore y(x) (x + 2)^6 = integral (x + 2)^6 (x + 2)^6 dx = integral (x + 2)^12 dx = (x + 2)^13/13 + c therefore y(x) = (x + 2)^7/13 + c(x + 2)^6 Now use y(0) = 6 y(0) = (0 + 2)^7/13 + c/(0 + 2)^6 = 2^7/ + c/6 \r\n therefore y(0) = 128/13 + c/64