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Mathematics 9 Online
OpenStudy (anonymous):

Find the specific solution of the following differential equation. Evaluate the solution x=.05

OpenStudy (anonymous):

\[dy/dx=-x/y\] \[y(0)=1\] \[y(0.5)=\]

OpenStudy (anonymous):

Solve the separable equation: \[\frac{ dy(x) }{ dx } = \frac{ -x }{ y(x) }\] Multiply both sides by y(x): \[\frac{ dy(x) }{ dx } y(x) = -x\] Integrate both sides with respect to x: \[\int\limits_{}^{} \frac{ dy(x) }{ dx } y(x) dx = \int\limits_{}^{} -x dx\] Evaluate the integrals (C is your constant): \[\frac{ y(x)^{2} }{ 2} = -\frac{ x ^{2} }{ 2 } + c\] Solve for y(x): \[y(x) = -\sqrt{-x ^{2}+2c}\]OR\[y(x) = \sqrt{-x ^{2}+2c}\] Simplify the arbitrary constants (because anything times c is still c): \[y(x) = -\sqrt{-x ^{2}+c}\]OR\[y(x) = \sqrt{-x ^{2}+c}\] Hope this is helpful! :D

OpenStudy (anonymous):

really really helpful thanks!

OpenStudy (anonymous):

I'm glad, good luck :D

OpenStudy (anonymous):

can you just explain to me really quick what happens when you take the square root

OpenStudy (anonymous):

Yea, so taking the square root is needed because you're solving for y(x) and it is squared. The reason there are two possible results is because when you take the square root of something it's usually +/- . For like y = x^2 - 4, if we set y equal to 0 and solve the answer is not 2, it is +/- 2.Then the constant just disappears because 2c is still a constant so it's still c.

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