- #1

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This is my work so far:

∫4y^3dy=∫x(x^2+1)dx

(y^4)/2=((x^2+1)^2)/2+c

The answer from the textbook is y=-(√(x^2+2)/2)

As you can see, my work will never equal the textbook answer when you put it in the y= stuff form. What did I do wrong?

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- Thread starter idks16
- Start date

- #1

- 1

- 0

This is my work so far:

∫4y^3dy=∫x(x^2+1)dx

(y^4)/2=((x^2+1)^2)/2+c

The answer from the textbook is y=-(√(x^2+2)/2)

As you can see, my work will never equal the textbook answer when you put it in the y= stuff form. What did I do wrong?

- #2

- 5

- 0

[itex] y(x) = -\sqrt{\frac{1}{2}(x^2+1)}[/itex]

and mathematica agrees with me, so perhaps a typo?

Anyway, it looks like your on the right track, although go back through the integration, I think you may be off by a factor.

Then apply the boundary condition to find the integration constant.

And simplify the algebra down to the answer.

Also be conscience of taking roots,

[itex] y(x) = ±(stuff)^{1/4}[/itex]

good luck

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