asilvester635
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Finding solution of the differential equation
- Thread starter asilvester635
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asilvester635
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Never mind I found the answer. The -4 should in the denominator should be -2. I found a simple arithmetic mistake while checking my calculations.
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The graphic is so small I'm not sure I'm reading it correctly. The following is my guess as to what you've posted:
\(\displaystyle \color{red}{(1)(a).\, \mbox{Find the solution.}}\)
\(\displaystyle \dfrac{dy}{dx}\, =\, xy^2,\, y(1)\, =\, 2\)
\(\displaystyle \int\, \dfrac{1}{y^2}\, dy\, =\, \int\, x\, dx\)
\(\displaystyle \color{blue}{\mbox{Integrate:}}\)
\(\displaystyle \int\, y^{-2}\, dy\, =\, \int\, x\, dx\)
\(\displaystyle -\dfrac{y^{-1}}{1}\, =\, \dfrac{x^2}{2}\, +\, C\)
\(\displaystyle \color{blue}{\mbox{Multiply both sides by }\, -1:}\)
\(\displaystyle y^{-1}\, =\, -\dfrac{x^2}{2}\, -\, C\)
\(\displaystyle y^{-1}\, =\, -\dfrac{x^{-2}}{2^{-1}}\, -\, C^{-1}\)
\(\displaystyle \color{blue}{\mbox{Multiply }\, C^{-1}\, \mbox{ by }\, \dfrac{2^{-1}}{2^{-1}}:}\)
\(\displaystyle y\, =\, -\dfrac{x^{-2}}{2^{-1}}\, -\, \dfrac{2^{-1}\, C^{-1}}{2^{-1}}\)
\(\displaystyle \color{blue}{\mbox{Combine:}}\)
\(\displaystyle y\, =\, -\dfrac{x^{-2}\, -\, 2^{-1}\, C^{-1}}{2^{-1}}\)
\(\displaystyle y\, =\, -\dfrac{2}{x^2\, -\, 2C}\)
\(\displaystyle \color{blue}{\mbox{Apply initial condition of }\, y(1)\, =\, 2:}\)
\(\displaystyle 2\, =\, -\dfrac{2}{(1)^2\, -\, 2C}\)
\(\displaystyle 2\, =\, -\dfrac{2}{1\, -\, 2C}\)
\(\displaystyle 2(1\, -\, 2C)\, =\, -2\)
\(\displaystyle 2\, -\, 2C\, =\, -2\)
\(\displaystyle 2(1\, -\, C)\, =\, -2\)
\(\displaystyle 1\, -\, C\,=\, -1\)
\(\displaystyle -C\, =\, -2\)
\(\displaystyle C\, =\, 2\)
\(\displaystyle \color{blue}{\mbox{Substitute }\, 2\, \mbox{ for }\, C:}\)
\(\displaystyle \color{green}{y\, =\, -\dfrac{2}{x^2\, -\, 4}}\)
Is the above a correct interpretation of your attachment? Thank you!
\(\displaystyle \color{red}{(1)(a).\, \mbox{Find the solution.}}\)
\(\displaystyle \dfrac{dy}{dx}\, =\, xy^2,\, y(1)\, =\, 2\)
\(\displaystyle \int\, \dfrac{1}{y^2}\, dy\, =\, \int\, x\, dx\)
\(\displaystyle \color{blue}{\mbox{Integrate:}}\)
\(\displaystyle \int\, y^{-2}\, dy\, =\, \int\, x\, dx\)
\(\displaystyle -\dfrac{y^{-1}}{1}\, =\, \dfrac{x^2}{2}\, +\, C\)
\(\displaystyle \color{blue}{\mbox{Multiply both sides by }\, -1:}\)
\(\displaystyle y^{-1}\, =\, -\dfrac{x^2}{2}\, -\, C\)
\(\displaystyle y^{-1}\, =\, -\dfrac{x^{-2}}{2^{-1}}\, -\, C^{-1}\)
\(\displaystyle \color{blue}{\mbox{Multiply }\, C^{-1}\, \mbox{ by }\, \dfrac{2^{-1}}{2^{-1}}:}\)
\(\displaystyle y\, =\, -\dfrac{x^{-2}}{2^{-1}}\, -\, \dfrac{2^{-1}\, C^{-1}}{2^{-1}}\)
\(\displaystyle \color{blue}{\mbox{Combine:}}\)
\(\displaystyle y\, =\, -\dfrac{x^{-2}\, -\, 2^{-1}\, C^{-1}}{2^{-1}}\)
\(\displaystyle y\, =\, -\dfrac{2}{x^2\, -\, 2C}\)
\(\displaystyle \color{blue}{\mbox{Apply initial condition of }\, y(1)\, =\, 2:}\)
\(\displaystyle 2\, =\, -\dfrac{2}{(1)^2\, -\, 2C}\)
\(\displaystyle 2\, =\, -\dfrac{2}{1\, -\, 2C}\)
\(\displaystyle 2(1\, -\, 2C)\, =\, -2\)
\(\displaystyle 2\, -\, 2C\, =\, -2\)
\(\displaystyle 2(1\, -\, C)\, =\, -2\)
\(\displaystyle 1\, -\, C\,=\, -1\)
\(\displaystyle -C\, =\, -2\)
\(\displaystyle C\, =\, 2\)
\(\displaystyle \color{blue}{\mbox{Substitute }\, 2\, \mbox{ for }\, C:}\)
\(\displaystyle \color{green}{y\, =\, -\dfrac{2}{x^2\, -\, 4}}\)
Is the above a correct interpretation of your attachment? Thank you!
asilvester635
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There's no need. I figured out the answer. Thanks though!
I'm with you down to
y-1 = -x2/2 - C
To continue though, I get
y-1 = - (x2 + C)/2
or
y = -2 / (x2 + C)
Can you continue from there?
D
Deleted member 4993
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yes, to the corner I go. I should have written a small c since C was an arbitrary constant 2C is just another arbitrary constant c.
D