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Angela H.

Calculus 1 / AB

2 months, 2 weeks ago

Okay, so in this question we have a given differential equation and we need to find his general solution. So, first of all, we need to write the auxiliary equation, so it is given by 7 r, squared plus 34 r minus 5, and that is equal to 0 point right. So here are middle term is minus 34 r. So we can write minus r, plus 35 right and the first and last terms are as it is right now for the first 2 terms, we can factor out our and for the last 2 terms, we can factor out hi right. So now we can factor out 7 r, minus 1 point right. So here r is equal to 1 upon 7 and r is equal to minus y right, which means r 1 is equal to 1 upon 7 and r 2 is equal to minus y point. So these are the 2 real root of the oxilary equation, right so for 2, real root. The general solution is given by c 1 multiplied by 1, just 2 r, 1 x, plus c 2 multiplied by us to 2 x. Right now substitute the value of r 1 and r 2 over here right. So this is the general solution right than.

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