Question

Find the general solution to the differential equation. y=eYx6 18. v OSCALC16.7.311. Find the definite integral. not Jo 3+2 19. OSCALC1 7.5.255. Use a table of integrals to evaluate the integral.

need help please!!!

Answer 1

Criven - y = ex eddy = x e gdy = xo da Integrating both sides fe ddy = fx da et = x + ştaking log} -V 26 (2+0) fylone) = ln (+) Any tb dt 3PQE 3tqt=u 2 dt = du dt=du

= f - du = [lace)] ( ln (3426)]. --=[ Int3+2) – 6543+240)] = {{In [5)- In (3)]. oifl.60 -7.0981 Jo 37 As 13eat - =.251) (19) - Banda 2 + 6 ---------------- het 22+6 = t. anda = dl- xdn = dt 3 dt Jo que 2 .

T t12dt Sternet 깨T 12tl Ja T) 삐 Replacing ty witha =3G-D 쉬3조 dn = OS87 o J제6

Answer 2

SOLUTION :

17.

y’ = e^(-y) x^6

=> e^y y’ = x^6 

=> ∫ e^y dy/dx * dx = ∫ x^6 dx

=>  e^y = x^7 / 7 + C

=> ln (e^y) = ln (x^7 / 7 + C) 

=> y = ln (x^7 / 7 + C)  (ANSWER).

 

18.

1

∫ dt / (3 + 2t) 

0

Let  3 + 2t = u

=> 2 dt = du

=> dt = du/2 

When t = 0, u = 3 and when t = 1, u = 5

                                 5

So, given integral = ∫    du / (2 u) 

                                 3

                     5

= 1/2 [ ln(u) ]

                     3

= 1/2 ( ln(5) - ln(3))

= 0.2554 (ANSWER).

19.

1

∫ 3x/(sqrt(x^2 + 8))  dx 

0

Let x^2 + 8 = u 

=> 2x dx = du 

=> 3/2 * 2x dx = 3/2 du 

=> 3x dx = 3/2 du

At x = 0, u = 8, at x = 1, u = 9

                               9

So given integral = ∫  3/2 u^(- 1/2) du

                               8

                              

                            9

= 3/2 [ 2 u ^(1/2) ] 

                            8

= 3 ( 9^(1/2) - 8^(1/2))

= 0.5147 (ANSWER).