Differential Equation: dy/dx = x/y^2 with y = 3 for x = 2

Becky4paws

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I'm improving...just not quite there on these:

Find the particular solution of the given differential equation the condtion of y=3 when x=2

dy/dx = x/y^2

y^2 d/dy = x d/dx
integral y^2 d/dy = integral x d/dx

=1/3y^3 + C = l/2 x^2 +C
y^3 = 3/2x^2 + 3C-sub3

The book gives the answer of y^3 = 3/2x^2 + 21...how did they come up with C = 21?
 
Re: Differential Equations

Hello, Becky4paws!"]I'm improving...just not quite there on these:

Find the particular solution of the given differential equation
with the condtion of y = 3 when x = 2

. . \(\displaystyle \frac{dy}{dx} \:= \:\frac{x}{y^2}\)

You are using strange notation . . .

Separate variables: \(\displaystyle \:\(\displaystyle y^2\,dy \:= \:x\,dx\)

Integrate" \(\displaystyle \int y^2\,dy \:= \:\int x\,dx\)

We have: \(\displaystyle \:\frac{1}{3}y^3 \:= \:\frac{1}{2}x^2\,+\,C\)

Then: \(\displaystyle \:y^3 \:= \:\frac{3}{2}x^2\,+\,C\)


We are given the initial condition: \(\displaystyle \,x\,=\,2,\:y\,=\,3\)

Plug them in: \(\displaystyle \:3^3\:=\:\frac{3}{2}(2^2)\,+\,C\;\;\Rightarrow\;\;27 \:=\:6\,+\,C\;\;\Rightarrow\;\;C\,=\,21\)

Therefore, the function is: \(\displaystyle \:y^3 \;=\;\frac{3}{2}x^2\,+\,21\)

\)
 
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