Find the particular solution of the differential equation dy + 7y = 6 satisfying the initial condition y(0) = 0. Answer: y= Your answer should be a function of x.

Answer:Step-by-step explanation:We have the equation [tex]y'+7y=6[/tex] with the initial condition [tex]y(0)=0[/tex]. It is not difficult to notice that this is a linear equation, which has the general expression[tex]y'+P(x)y=Q(x)[/tex].The solution of this equation is expressed by a general formula:[tex]y(x) = \exp\left(-\int P(x)dx\right)\left(\int Q(x)\exp\left(-\int P(x)dx\right) +C\right)[/tex].In the particular case of our equation, we have[tex]P(x)=7[/tex][tex]Q(x)=6[/tex].Then, we must calculate the integrals[tex]\int 7dx = 7x[/tex] that implies[tex]\exp\left(-\int P(x)dx\right) = e^{-7x}[/tex],and[tex]\int 6e^{7x}dx = \frac{6}{7}e^{7x}[/tex]Then,[tex]y(x) = e^{-7x}\left(\frac{6}{7}e^{7x} +C\right) = \frac{6}{7} + Ce^{-7x}[/tex].In order to obtain the value of the constant we substitute the initial condition[tex] 0=y(0) = \frac{6}{7} + C[/tex] that implies [tex]C=-\frac{6}{7}[/tex]Therefore,[tex]y(x) = \frac{6}{7}-\frac{6}{7}e^{-7x}[/tex].