1 Problem 1 Give the general solution y0 = x2/y ydy = x2 dx ⇒ 1 2 y2 = 1 3 x3 C 2 Problem 3 Give the general solution to y0 y2 sin(x) = 0 First write in standard form dy dx = −y2 sin(x) ⇒ − 1 y2 dy = sin(x)dx Before going any further, notice that we have divided by y, so we need to say that this is value as long as y(x) 6= 0Find dy/dx y^2=(x1)/(x1) Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate using the chain rule, which states that is where and Tap for more steps To apply the Chain Rule, set asAnswer (1 of 2) homogeneous part is y''y'y=0 which is constant coefficient, so assume y=exp(m*x) then you get m^2m1=0 so m=(1 pm sqrt(1–4*1*1))/(2*1)= (1 pm sqrt(3)*%i)/2 where pm is plus or minus Since the inhomogeneous term does not include exp(m*x), we can look for a particular soluti Engineering Mathematics Notes (x+1)dy/dx-y=e^3x(x+1)^2
