twoflower
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Hi all,
I'm given this one:
<br /> y' = 10^{x+y}<br />
Here's how I went:
<br /> \frac{y'}{10^{y}} = 10^{x}<br />
<br /> \int\left(\frac{1}{10}\right)^{y}\ dy = \int 10^{x}\ dx<br />
<br /> \frac{\left(\frac{1}{10}\right)^{y}}{\log \frac{1}{10}}} = \frac{10^{x}}{\log 10} + C<br />
<br /> \left(\frac{1}{10}\right)^{y} = -10^{x}<br />
From which I conclude that it has no solution (on the right side there is negative number whereas on the left side there always will be a positive one).
Is this right conclusion?
Thank you.
I'm given this one:
<br /> y' = 10^{x+y}<br />
Here's how I went:
<br /> \frac{y'}{10^{y}} = 10^{x}<br />
<br /> \int\left(\frac{1}{10}\right)^{y}\ dy = \int 10^{x}\ dx<br />
<br /> \frac{\left(\frac{1}{10}\right)^{y}}{\log \frac{1}{10}}} = \frac{10^{x}}{\log 10} + C<br />
<br /> \left(\frac{1}{10}\right)^{y} = -10^{x}<br />
From which I conclude that it has no solution (on the right side there is negative number whereas on the left side there always will be a positive one).
Is this right conclusion?
Thank you.
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