Solve the given differential equation by separation of...

Transcript

00:01 We have been given a differential equation dp d t equal to p minus p square we have to use separation of variables to to different to solve this equation so what we'll do we will take p on the one side of the equation and t on the another side of the equation so we will divide we will multiply both sides by d t so this becomes p minus p square d t d p is equal to and we will divide both sides by p minus p square so this is dp by p minus b square equal to d t and now we will arrange this denominator as the difference of two square terms or two square expressions for that we will add and subtract 1 by 4 so this is equal to d t 1 by 4 can be written as half square and subtract factoring out minus 1 we get here p square minus p plus 1 by 4 is equal to d t therefore this is dp by half square minus it can be written as p minus half square is equal to d t now to integrate a function of this form we will use the integration formula dx by a square minus x square is equal to 1 by twice a log x plus or a plus x by a minus x plus an integrating constant c so using this equation this integration formula we will integrate both sides of this equation as a here and x is considered as p minus half so we will get log half plus p minus half minus half is equal to t plus an integrating constant let us consider lock c so this will become one and this will become half and minus up will cancel out so this is lock p by 1 minus p is equal to t plus lock c t plus log now we will raise both sides of this equation to the power of e that is e to the power log p by 1 minus p equal to a to the power t plus lox see now e to the power log x is actually x so this will be become p equal to p by 1 minus p and it can be written as a to the part t and u to the power lox c e to the bar lock c is equal to c.

03:15 So it can be written as p by 1 minus p is equal to c to the board t now we will cross multiply this 1 minus p on the right hand side.

03:28 So if this becomes c to the bot t 1 minus p therefore p is equal to c e to the bot t minus c p e to the bot t and we will take this on the left hand side so this becomes p plus c p e to the bar t is equal to c to the per t...

Question

Solve the given differential equation by separation of variables. dP/dt = P - P^2

Question

Solve the given differential equation by separation of variables. dP/dt = P - P^2

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