Lynn_gonsalves
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- Apr 21, 2021
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- 1
Can someone give me the answer of this
- Thread starter Lynn_gonsalves
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Dr.Peterson
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- Nov 12, 2017
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That's not what we do here:
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Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...
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Please show us where you need help, so we can assist you in learning to solve it.
And tell us why you need the answer, when it appears that what comes next is the answer.
To get you started, write [MATH]\sin2y \text{ as }2\sin y\cos y[/MATH]
[MATH]\frac{dy}{dx}=x^3 \cos^2 y- 2x\sin y \cos y \hspace7ex \text{(1)}[/MATH]
and then you could try a substitution [MATH]\boxed{\hspace1ex u=\tan y \hspace1ex}[/MATH]
so, [MATH]\frac{du}{dx}=\sec^2y \frac{dy}{dx}= \hspace1ex ...[/MATH]and complete this using the expression for [MATH]\frac{dy}{dx}[/MATH] in equation (1) above.
This should give you something you can solve.
[MATH]\frac{dy}{dx}=x^3 \cos^2 y- 2x\sin y \cos y \hspace7ex \text{(1)}[/MATH]
and then you could try a substitution [MATH]\boxed{\hspace1ex u=\tan y \hspace1ex}[/MATH]
so, [MATH]\frac{du}{dx}=\sec^2y \frac{dy}{dx}= \hspace1ex ...[/MATH]and complete this using the expression for [MATH]\frac{dy}{dx}[/MATH] in equation (1) above.
This should give you something you can solve.
Last edited:
Actually I don't know much about differential equations. I just got lucky! But thanks.
So true. In fact I thought about saying 'practice' rather than 'luck'.
What do we conclude then from:
(1) "Practice makes you luckier", and
(2) Eisenhower (and Napoleon before him): “I'd rather have a lucky general than a smart general"?
Indeed. Megalomaniacs tend to push their luck.