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timofeeve [1]
4 years ago
14

(1 point) solve the separable differential equation y′(x)=2y(x)+18−−−−−−−−−√, y′(x)=2y(x)+18, and find the particular solution s

atisfying the initial condition y(5)=9.
Mathematics
1 answer:
luda_lava [24]4 years ago
7 0
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
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