Question

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 1

Answer:

Step-by-step explanation:

We have the equation $y'+7y=6$ with the initial condition $y(0)=0$. It is not difficult to notice that this is a linear equation, which has the general expression

$y'+P(x)y=Q(x)$.

The solution of this equation is expressed by a general formula:

$y(x) = \exp\left(-\int P(x)dx\right)\left(\int Q(x)\exp\left(-\int P(x)dx\right) +C\right)$.

In the particular case of our equation, we have

$P(x)=7$

$Q(x)=6$.

Then, we must calculate the integrals

$\int 7dx = 7x$ that implies

$\exp\left(-\int P(x)dx\right) = e^{-7x}$,

and

$\int 6e^{7x}dx = \frac{6}{7}e^{7x}$

Then,

$y(x) = e^{-7x}\left(\frac{6}{7}e^{7x} +C\right) = \frac{6}{7} + Ce^{-7x}$.

In order to obtain the value of the constant we substitute the initial condition

$0=y(0) = \frac{6}{7} + C$ that implies $C=-\frac{6}{7}$

Therefore,

$y(x) = \frac{6}{7}-\frac{6}{7}e^{-7x}$.