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Numerade Educator



Problem 34 Hard Difficulty

An integral equation is an equation that contains an unknown function $ y(x) $ and an integral that involves $ y(x). $ Solve the given integral equation. [Hint: Use an initial condition obtained from the integral equation.]
$ y(x) = 2 + \int^x_1 \frac {dt}{ty(t)}, x > 0 $


$y^{2}=2 \ln x+4$


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Video Transcript

this question here asked us to solve the given integral equation as they have given us wife acts. So what we know we have is we have wives. Axe is two plus bounds of one acts d t over t y of tea. Remember, we're looking specifically the variable of teeth. What we know is we need to write this differentiated in terms of d y de axe as one over X y of acts again, we're differentiating with respect to X. This is equivalent to why d why is d axe over X? This means that why squared over two is not your axe pussy. Remember your rules for like, one over acts and then looking at the differential of this given Ln our natural log and its relationship to different differentiating Now you can substitute Why have one or X is essentially being substituted for one and we get why of one is too and then you know when you have the same bound upper and lower, this is just zero. So now that we have X is one and wise too, we know we're plugging in X is one is wise to so two squared over two is naturally one plus. See, you can also use Kay. Remember not one of one of the same thing A zero, which means to is C because two is zero plus e substitute back. Because remember, we're solving for see because then we can write our original value. And we have why squared is to natural other acts. Plus four we got the plus four by multiplied by two to get rid of this 200 denominator.