Respuesta :
Brief review of proportionality relationships:
When two quantities [tex]a,b[/tex] are [tex]\textbf{directly}[/tex] proportional, that means any change in [tex]a[/tex] manifests a [tex]\textbf{direct}[/tex] change (think "in the same direction") in [tex]b[/tex].
Silly example: "The more I eat, the fatter I get." Here the amount one eats is directly proportional to one's body weight.
This change isn't always one-for-one, so we introduce a constant [tex]k[/tex] to account for any scaling that occurs on either variables behalf. In general, though, we can write a directly proportional relationship as [tex]a=kb[/tex].
Now, when [tex]a,b[/tex] are [tex]\textbf{inversely}[/tex] proportional, then a change in [tex]a[/tex] manifests a change in [tex]b[/tex] in the [tex]\textbf{inverse}[/tex] (opposite) direction.
Silly example: "The more I eat, the less thin I get."
This time we write the relation as [tex]ab=k[/tex].
To get back to your problem: To say that the rate of change of [tex]y(x)[/tex] is inversely proportional to [tex]\sqrt y[/tex] is to say that there is some constant [tex]k[/tex] such that
[tex]\sqrt y\dfrac{\mathrm dy}{\mathrm dx}=k[/tex]
This is a separable ODE:
[tex]y^{1/2}\,\mathrm dy=k\,\mathrm dx[/tex]
[tex]\displaystyle\int y^{1/2}\,\mathrm dy=\int k\,\mathrm dx[/tex]
[tex]\dfrac23y^{3/2}+C_y=kx+C_x[/tex]
[tex]\dfrac23y^{3/2}=kx+C[/tex]
[tex]y^{3/2}=\dfrac{3k}2x+C[/tex]
[tex]y=\left(\dfrac{3k}2x+C\right)^{2/3}[/tex]
When two quantities [tex]a,b[/tex] are [tex]\textbf{directly}[/tex] proportional, that means any change in [tex]a[/tex] manifests a [tex]\textbf{direct}[/tex] change (think "in the same direction") in [tex]b[/tex].
Silly example: "The more I eat, the fatter I get." Here the amount one eats is directly proportional to one's body weight.
This change isn't always one-for-one, so we introduce a constant [tex]k[/tex] to account for any scaling that occurs on either variables behalf. In general, though, we can write a directly proportional relationship as [tex]a=kb[/tex].
Now, when [tex]a,b[/tex] are [tex]\textbf{inversely}[/tex] proportional, then a change in [tex]a[/tex] manifests a change in [tex]b[/tex] in the [tex]\textbf{inverse}[/tex] (opposite) direction.
Silly example: "The more I eat, the less thin I get."
This time we write the relation as [tex]ab=k[/tex].
To get back to your problem: To say that the rate of change of [tex]y(x)[/tex] is inversely proportional to [tex]\sqrt y[/tex] is to say that there is some constant [tex]k[/tex] such that
[tex]\sqrt y\dfrac{\mathrm dy}{\mathrm dx}=k[/tex]
This is a separable ODE:
[tex]y^{1/2}\,\mathrm dy=k\,\mathrm dx[/tex]
[tex]\displaystyle\int y^{1/2}\,\mathrm dy=\int k\,\mathrm dx[/tex]
[tex]\dfrac23y^{3/2}+C_y=kx+C_x[/tex]
[tex]\dfrac23y^{3/2}=kx+C[/tex]
[tex]y^{3/2}=\dfrac{3k}2x+C[/tex]
[tex]y=\left(\dfrac{3k}2x+C\right)^{2/3}[/tex]
Applying separation of variables, the solution to the differential equation [tex]\frac{dy}{dx} = \frac{1}{\sqrt{y}}[/tex] is:
[tex]y(x) = \sqrt[3]{(1.5x + K)^2}[/tex]
"The rate of change of y with respect to x is inversely proportional to the square root of y", hence, the differential equation is:
[tex]\frac{dy}{dx} = \frac{1}{\sqrt{y}}[/tex]
What is separation of variables?
In separation of variables, we place all the factors of y on one side of the equation with dy, all the factors of x on the other side with dx, and integrate both sides.
Hence:
[tex]\frac{dy}{dx} = \frac{1}{\sqrt{y}}[/tex]
[tex]\sqrt{y} dy = dx[/tex]
[tex]\int y^{\frac{1}{2}}dy = dx[/tex]
[tex]\frac{2y^{\frac{3}{2}}}{3} = x + K[/tex]
In which K is the constant of integration.
[tex]y^{\frac{3}{2}} = 1.5x + K[/tex]
[tex]y(x) = \sqrt[3]{(1.5x + K)^2}[/tex]
To learn more about separation of variables, you can take a look at https://brainly.com/question/14318343
