+0  
 
0
303
13
avatar+291 

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 :)

lakofbrwn  Feb 8, 2016

Best Answer 

 #4
avatar+26329 
+10

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

 

equality

Alan  Feb 8, 2016
Sort: 

6+0 Answers

 #1
avatar
0

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+26329 
+10
Best Answer

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

 

equality

Alan  Feb 8, 2016
 #10
avatar
0

.....................

Guest Feb 8, 2016
 #11
avatar+91039 
0

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

Melody  Feb 9, 2016
 #12
avatar+18715 
+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

heureka  Feb 9, 2016
 #13
avatar
0

if

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

then, differentiating,

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

Guest Feb 9, 2016

20 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details