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# a good way to jump start your brain for the mathletes

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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 :)

lakofbrwn  Feb 8, 2016

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

Alan  Feb 8, 2016
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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
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Here's the same solution as Guest's, laid out using mathematical notation:

Alan  Feb 8, 2016
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.....................

Guest Feb 8, 2016
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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.

Melody  Feb 9, 2016
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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 $$y'=2xy$$

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

heureka  Feb 9, 2016
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if

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

then, differentiating,

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

Guest Feb 9, 2016