Help with a Partial derivative problem

Could someone please help me with this problem? Thanks.

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Sure. What help do you need? Have you at least started, perhaps by integrating each of the derivatives?

Please follow our guidelines, which you should have read before starting; here is the quick version:

Note that it's particularly helpful to tell us your context, so we know whether you are studying differential equations, or just learned partial derivatives. I suspect the latter. But the main thing we ask for is some work.
 
Thanks Dr. Peterson.
I have tried to integrated both the derivatives, but I am not sure whether the answer I get is correct or not.
Thanks again for your great help.
 

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Here is my approach. Please could you help me check my answer?

[MATH]\because f_x(x,y)=4e^{4x}y^2[/MATH] and [MATH]f_y(x,y)=2e^{4e}y[/MATH]
[MATH]\therefore f(x,y)=e^{4x}y^2+C[/MATH], where [MATH]C[/MATH] is a constant.

Let [MATH]K[/MATH] be a constant s.t. [MATH]f(x,y)=K[/MATH] is the mentioned iso curve.

[MATH]\because[/MATH] the iso curve passes through [MATH](0,2)[/MATH]
[MATH]\therefore f(0,2)=e^{4\times0}(2^2)+C=k[/MATH]
[MATH]\therefore 4+C=k[/MATH]
Recall that the iso curve is [MATH]f(x,y)=k[/MATH],

therefore, [MATH]f(x,y)=k\Rightarrow e^{4x}y^2+C=k\Rightarrow e^{4x}y^2+C=4+C\Rightarrow e^{4x}y^2=4\Rightarrow y=\frac{2}{e^{2x}}[/MATH]
Is [MATH]y=\frac{2}{e^{2x}}[/MATH] the analytical form of [MATH]g(x)[/MATH]?

Many thanks.
 
Last edited:
Here is my approach. Please could you help me check my answer?

[MATH]\because f_x(x,y)=4e^{4x}y^2[/MATH] and [MATH]f_y(x,y)=2e^{4e}y[/MATH]
[MATH]\therefore f(x,y)=e^{4x}y^2+C[/MATH], where [MATH]C[/MATH] is a constant.

Let [MATH]K[/MATH] be a constant s.t. [MATH]f(x,y)=K[/MATH] is the mentioned iso curve.

[MATH]\because[/MATH] the iso curve passes through [MATH](0,2)[/MATH]
[MATH]\therefore f(0,2)=e^{4\times0}(2^2)+C=k[/MATH]
[MATH]\therefore 4+C=k[/MATH]
Recall that the iso curve is [MATH]f(x,y)=k[/MATH],

therefore, [MATH]f(x,y)=k\Rightarrow e^{4x}y^2+C=k\Rightarrow e^{4x}y^2+C=4+C\Rightarrow e^{4x}y^2=4\Rightarrow y=\frac{2}{e^{2x}}[/MATH]
Is [MATH]y=\frac{2}{e^{2x}}[/MATH] the analytical form of [MATH]g(x)[/MATH]?

Many thanks.
Yes, that's good. You fixed the error in the first attempt.
 
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