Shown above is a slope field for the differential equation (...

Shown above is a slope field for the differential equation ( frac{d y}{d x}=y^{2}left(4-y^{2} ight) ). If ( y=g(x) ) is the solution to the differential equation with the initial condition ( g(-2)=-1 ), then, ( lim _{x ightarrow infty} g(x) ) is (A) ( -infty ) (B) ( -2 ) (C) 0 (D) 2 (E) 3

Transcript

00:01 Hello everyone, in this question the differential equation is given we have to find the limit x tends to infinity g of x.

00:07 So first we can write by using separation of variable dy by y square 4 minus y square equal to dx.

00:17 So integrating on both sides integral of dy y square minus 4 y square integral of dx we will be having integral of 1 by 4 this can be written as 1 by 4 4 minus y square plus y square divided by y square 4 minus y square dy integral of dx.

00:43 So this will become 1 by 4 4 minus y square plus y square divided by y square 4 minus y square dy as 1 by 4 integral of 4 minus y square divided by y square 4 minus y square dy plus integral of y square by y square 4 minus y square dy.

01:13 That is i have separated this two this is there now i am solving only the lhs part.

01:21 So this again become this 4 minus y square by 4 minus y square will become 1 by 4 integral of 1 by y square canceling we will be having integral of 1 by 4 minus y square dy.

01:42 So this can be written as 1 by 4 integral of 1 by y square dy this common actually so plus this 1 by 4 square minus y square can be written as 1 by 2 square minus y square dy.

01:58 So this will become so on integrating we will be having 1 by 4 minus 1 by y plus 1 by 2 into 2 log of 2 plus y divided by 2 minus y.

02:12 So this this is actually lhs rhs part is we know integral of dx.

02:19 So integral of dx is nothing but x plus c.

02:22 So now comparing lhs equal to rhs we will be having integral of dy that is y square into 4 minus y square equal to dx.

02:37 So this only we have found that is 1 by 4 minus 1 by y plus 1 into 2 into 2 log of 2 plus y 2 minus y x plus c.

02:48 So this 4 will come to this side so minus 1 by y plus 1 by 2 into 2 log of 2 plus y 2 minus y 4x plus c.

03:00 So we have found this equation.

03:03 Now we have to apply that is y equal to g of x and we know g of minus 2 equal to minus 1.

03:11 So here x is minus 2 and y is minus 1.

03:16 So we have to apply this value here.

03:18 So we'll be having so putting x and y value in the above step we'll be having minus 1 by y plus 1 by 2 into 2 log of 2 plus y 2 minus y.

03:30 This is the required equation where we put x as minus 2 and y as minus 1.

03:35 We will be having minus 1 by minus 1 plus 1 by 2 into 2 log of 2 minus 1 2 plus 1 4 into minus 2 plus c...

Question

Shown above is a slope field for the differential equation ( frac{d y}{d x}=y^{2}left(4-y^{2} ight) ). If ( y=g(x) ) is the solution to the differential equation with the initial condition ( g(-2)=-1 ), then, ( lim _{x ightarrow infty} g(x) ) is (A) ( -infty ) (B) ( -2 ) (C) 0 (D) 2 (E) 3

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