We write the equation in terms of dy/dx,
<span>y'(x)=sqrt (2y(x)+18)</span>
dy/dx = sqrt(2y + 18)
dy/dx = sqrt(2) ( sqrt(y + 9))
Separating the variables in the equation, we will have:
<span>1/sqrt(y + 9) dy= sqrt(2) dx </span>
Integrating both sides, we will obtain
<span>2sqrt(y+9) = x(sqrt(2)) + c </span>
<span>where c is a constant and can be determined by using the boundary condition given </span>
<span>y(5)=9 : x = 5, y = 9
</span><span>sqrt(9+9) = 5/sqrt(2) + C </span>
<span>C = sqrt(18) - 5/sqrt(2) = sqrt(2) / 2</span>
Substituting to the original equation,
sqrt(y+9) = x/sqrt(2) + sqrt(2) / 2
<span>sqrt(y+9) = (2x + 2) / 2sqrt(2)
</span>
Squaring both sides, we will obtain,
<span>y + 9 = ((2x+2)^2) / 8</span>
y = ((2x+2)^2) / 8 - 9
A lot of numbers and it really depends
X + 5 + 11x = 12x + y
Simplify: 12x + 5 = 12x + y (We are adding x and 11x on the left side)
Subtract 12x from each side makes each zero.
5 = y
So we can plug in and test. I'm picking two random numbers to plug in for x. 10, and 82
x = 10, y = 5
10 + 5 + 11(10) = 12(10) + 5
125 = 125
x = 82, y = 5
82 + 5 + 11(82) = 12(82) + 5
989 = 989
So we verified y = 5