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Good morning, was wondering if anyone can figure this out, i'll post the answer later 

Verify that y=5ex^2 is a solution of y'=2xy

 

Good Luck :)

 Feb 8, 2016

Best Answer 

 #4
avatar+28182 
+10

Here's the same solution as Guest's, laid out using mathematical notation:

 

equality

 Feb 8, 2016
 #1
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Solve the separable equation ( dy(x))/( dx) = 2 x y(x):                                                                                Divide both sides by y(x): (( dy(x))/( dx))/(y(x)) = 2 x Integrate both sides with respect to x: integral (( dy(x))/( dx))/(y(x)) dx = integral 2 x dx Evaluate the integrals: log(y(x)) = x^2+c_1, where c_1 is an arbitrary constant. Solve for y(x): y(x) = e^(x^2+c_1) Simplify the arbitrary constants:                  Answer: | | y(x) = c_1 e^(x^2)

Guest Feb 8, 2016
 #4
avatar+28182 
+10
Best Answer

Here's the same solution as Guest's, laid out using mathematical notation:

 

equality

Alan Feb 8, 2016
 #10
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.....................

 Feb 8, 2016
 #11
avatar+105604 
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Thanks Alan and lakofbryn :)

It is easy to show this but as a mental exercise I really like what Alan did.

It took me a little while to understand it though.    laugh

 Feb 9, 2016
 #12
avatar+23299 
+5

Good morning, was wondering if anyone can figure this out, i'll post the answer later 

Verify that \(y=5e^{x^2}\) is a solution of \(y'=2xy\)

 

\(\begin{array}{rcll} y &=& 5e^{x^2} \qquad & \ln{()} \\ \ln{(y)} &=& \ln{( 5e^{x^2} ) } \\ \ln{(y)} &=& \ln{( 5 ) } + \ln{( e^{x^2} ) } \\ \ln{(y)} &=& \ln{( 5 ) } + x^2\cdot \ln{( e ) } \qquad & \ln{(e)} = 1 \\ \ln{(y)} &=& \ln{( 5 ) } + x^2 \qquad & \frac{d()}{dx} \\ \frac{y'}{y} &=& 0 + 2\cdot x \\ \frac{y'}{y} &=& 2\cdot x \\ y' &=& 2\cdot x \cdot y \\ \end{array}\)

 

laugh

 Feb 9, 2016
 #13
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if

\(\displaystyle y = 5e^{x^{2}} ,\)

then, differentiating,

\(\displaystyle y ' = 5e^{x^{2}}(2x) = 2xy.\)

.
 Feb 9, 2016

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