Question 8 Solve the IVP √(y)dx + (x

Question 8 Solve the IVP $\sqrt{y}dx + (x - 4)dy = 0$, $y(5) = 49$ $-2\sqrt{y} = \ln|x - 4| - 14$ $-2y^{\frac{3}{2}} = \ln|x - 4| - 14$ $-2y^{\frac{3}{2}} = \ln|x - 4| + 14$ $-2\sqrt{y} = \ln|x - 4| + 12$ 0.75 pts

Transcript

00:01 Hi, so first it is 1 .1.

00:05 Solve the differential equations i .e.

00:09 D square y of t divided by dt square plus 4y of t equals to 8t square minus 20t plus 8.

00:22 We will now find the complementary solution to the lhs part equals to 0.

00:31 We say that let y of t be e raise to lambda t d square by dt square of e raise to lambda t plus 4 e raise to lambda t equals to 0.

00:51 So that means lambda square e raise to lambda t plus 4 e raise to lambda t equals to 0.

01:01 So that means this implies that lambda is equal to plus minus 2i.

01:09 Y of t will be equal to c1 e raise to 2it plus c2 e raise to minus 2it and so we can say that y of t will be c1 of cos of 2t plus i sin of 2t plus c2 cos of 2t minus i sin of 2t.

01:46 So y of t will be c1 plus c2 cos of 2t plus i c1 minus c2 sin of 2t.

02:01 So now we will be finding the particular solution yp of t will be a1 plus a2t plus a3t square.

02:23 So first now we will find yp double dash and we get it as 2a cube yp dash it is a2 plus 2a cube t.

02:43 Further solving we get a1 as 1 a2 as minus 5 a3 as 2.

03:02 So y of t will be 2t square minus 5t plus c1 cos of 2t plus c2 sin of 2t plus 1.

03:18 So here we have second one y double dash of t plus y dash of t minus 2y of t equals to 3 cos of 2t.

03:28 Also the initial conditions given that is y of 0 is minus 1 y dash of 0 is 2.

03:34 We find the complementary solution to the left -hand side...

Question

Question 8 Solve the IVP $\sqrt{y}dx + (x - 4)dy = 0$, $y(5) = 49$ $-2\sqrt{y} = \ln|x - 4| - 14$ $-2y^{\frac{3}{2}} = \ln|x - 4| - 14$ $-2y^{\frac{3}{2}} = \ln|x - 4| + 14$ $-2\sqrt{y} = \ln|x - 4| + 12$ 0.75 pts

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