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

y(x) = (6/7)*e^(7*x) - (6/7)

Step-by-step explanation:

I will use the notation:

dy/dx = y'

We have:

y' + 7*y = 6.

Because in the right side we do not have any term that depends on x, we can assume that y is an exponential function:

y = a*e^(b*x) + c

where a, b and c are constants.

y' = b*a*e^(b*x)

First, let's apply the condition in zero.

y(0) = 0 = a*e^(b*0) + c = a + c = 0.

then we have c = -a.

And we can write our equation as:

y = a*e^(b*x) - a.

Then the differential equation becomes:

b*a*e^(b*x) - 7*a*e^(b*x) + 7*a = 6.

In the left side we do not have any exponential, then we must have b = 7 (So the exponentials cancel eachother)

7*a*e^(7*x) - 7*a*e^(7*x) + 7*a = 6

7*a = 6

a = 6/7.

Our equation is:

y(x) = (6/7)*e^(7*x) - (6/7)

a. -0.3 > -0.33

quit

step-by-step explanation: