y' + 2xy = 10x Solve the first-order differential equation

y'+2xy =10x
To solve, re-write the derivative as dy/dx .
dy/dx + 2xy = 10x
Then, bring together same variables on one side of the equation.
dy/dx = 10x - 2xy
dy/dx = 2x(5 - y)
dy/(5-y) = 2x dx
Next, take the integral of both sides.
int dy/(5-y) = int 2xdx
-ln |5-y| +C_1= (2x^2)/2 + C_2
Then, isolate the y.
-ln|5-y| = x^2+C_2-C_1
ln|5-y|=-x^2- C_2 +C_1
Since C1 and C2 represent any number, express it as a single constant C.
ln|5-y| = -x^2+ C
e^(ln|5-y|) = e^(-x^2+C)
|5-y| = e^(-x^2+C)
5-y = +-e^(-x^2+C)
Applying the exponent rule a^m*a^n = a^(m+n) ,
the right side becomes
5-y = +- e^(-x^2)*e^C
5-y = +-e^C*e^(-x^2)
-y = +-e^C*e^(-x^2) - 5
y = +-e^C*e^(-x^2)+5
Since+-e^C is a constant, it can be replaced by a constant C.
y = Ce^(-x^2) + 5
 
Therefore, the general solution is  y = Ce^(-x^2) + 5 .