(2+x)y' = 3y Find the general solution of the differential equation
Recall that y' is the same as (dy)/(dx) . Then in the given problem: (2+x)y'=3y , we may write it as: (2+x) (dy)/(dx) = 3y. This will help to follow the variable separable differential equation in a form of N(y) dy = M(x) dx. To rearrange (2+x) (dy)/(dx) = 3y ,cross-multiply (dx) to the other side: (2+x)dy =3y dx Divide both sides by (2+x) : ((2+x)dy)/(2+x) =(3y dx)/(2+x) dy =(3y dx)/(2+x) Divide both sides by y : (dy )/y=(3y dx)/((2+x)y) (dy)/y=(3dx)/(2+x) To solve for the general solution of the differential equation, apply direct integration on both sides: int (dy)/y=int (3dx)/(2+x) For the left side, apply the basic integration formula for logarithm: int (dy)/y= ln|y| For the right side, we may apply the basic integration property: int c*f(x) dx = c int f(x)dx . int (3dx)/(2+x)= 3 int (dx)/(2+x) Let u =2+x then du= dx The integral becomes: 3 int (dx)/(2+x) = 3 int (du)/u We can now apply the basic integration formula for logarithm on the integral part: 3 int (du)/u= 3l...