Question

A first order linear equation in the form y′+p(x)y=f(x) y p x y f x can be solved by finding an integrating factor μ(x)=exp(∫p(x)dx) μ x exp p x d x (1) Given the equation y′+6y=4 y 6 y 4 find μ(x)= μ x (2) Then find an explicit general solution with arbitrary constant C C . y= y . (3) Then solve the initial value problem with y(0)=3 y 0 3 y= y .

Answer 1

Given differential equatio is $y'+6y=4$ , Here p(x) =6

Hence the integrating factor is $\mu =e^{\int 6dx}=e^{6x}$

The general solution to the given differential equation is then given as

$y(x)=\frac{1}{e^{6x}}\left (\int (4)e^{6x}dx+C \right ) \\ ~~~~~~~~~~~~=\frac{2}{3}+Ce^{-6x}$

As y(0)= 3 so we have

$3=\frac{2}{3}+C ~,~C=\frac{7}{3}$

Hence the solution required is $y(x)=\frac{2}{3}+\frac{7}{3}e^{-6x}$ or $3y(x)=2+7e^{-6x}$