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
Answer:
First odd number = 13
Second odd number = 15
Step-by-step explanation:
Let
First odd number = 2x + 1
Second odd number = 2x + 3
According to given conditions:
32 + 2x + 1= 3 (2x + 3)
33 + 2x = 6x + 9
Taking terms with x to left and constants to right
2x - 6x = 9 - 33 (sign of transferred terms will be changed)
-4x = -24
Dividing both sides by -4
x = -24/-4
x =6
So,
First odd number = 2x + 1 = 2*6 + 1 = 12 + 1 =13
Second odd number = 2x + 3= 2*6 +3= 12 + 3 = 15
I hope it will help you!
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