Comparing with M dx + N dy = 0
M = a y^3 e^(axy) + b
N = 2 y e^(axy) + a x y^2 e^(axy) - 1
Partial derivative of M wrt y ,
M[y] = 3 a y^2 e^(axy) + a^2 x y^3 e^(axy)
Partial derivative of N wrt x ,
N[x] = 2 a y^2 e^(axy) + a y^2 e^(axy) + a^2 x y^3 e^(axy)
=> N[x] = 3 a y^2 e^(axy) + a^2 x y^3 e^(axy)
Clearly , M[y] = N[x]
Exact differential equation.
Solution is
y^2 e^(axy) + b x - y = c
Answer --> (d)
Hope this helps.
Thanks.
Comment if you have any questions.
If a and b are constants, the solution of (ayaw+b)dx +(2ye KY + axy W -...
If a and b are constants, the solution of (aye axy +b)dx + (2ye axy + axy e axy – 1)dy=0 is Select one: a.y? (e axy – 1)=c b. ye ax + bx-y=c c. ye axy + x-y=C d. ye axy + bx=y e. ye axy + bx-y=C f. eaxy + bx = C g. y2e axy + bx+y=C h.yze axy + bx-y=c
If a and b are constants, the solution of (aye « +b)dx +(2ye xy + axy e ay – 1)dy=0 is a. Select one: ..y?e@y+bx = y b.y?(exy - 1) = 0 o c.yle axy +bx+y=C O d. ye xy + bx-y=C oe.yle ax + bx-y=C of.yle xy +bx-y=C g. e ary+bx= c o h. ye xy + x-y=C
If a and b are constants, the solution of (ayže xy +b)dx + (2ye xy + axy e xy – 1)dy=0 is Select one: oa.yle xy + x-y=C O b.yle xy +bx+y=C O c. y2exy – 1)=C O d. ye xy + bx-y=C O e.yle xy + bx=y f. e xy + bx=C O g. yle" + bx-y=C O h. ye xy + bx-y=C
IVI U OT 6.00 p Flag question If a and b are constants, the solution of (aye axy+b)dx+(2ye xy + axy’e x – 1)dy=0 is Select one: a.yle axy + bx-y=c b. eaxy + bx= c O c. ye axy + bx - y = c d. y2e xy + x - y=c e. ye ax + bx-y=C f. ye xy +bx+y=c g. y? (e axy - 1) = 0 h. Y x + bx = y
Marked out of 6.00 Pidg queJLIUM Question 3 Not yet answered if a and b are constants, the solution of (aye axy +b)dx+(2ye axy + axy e axy – 1)dy=0 is Select one: a. ewy + bx=C o bye + bx-y= 0 O c. ye xy + bx-y=C odyexy +bx+y=C oe.y?e@y+bx = y o .yeasy + x-y=C o & ylemy -1)=C Oh.ye® + bx-y=C
Question 5 Not yet answered Marked out of 6.00 p Flag question If a and b are constants, the solution of (ayže xy +b)dx + 2ye xy + axy e x – 1 )dy=0 is Select one: O a.yle + bx-y=C O b. ye wy+x-y=C O ce@xy + bx=0 O d. ye xy + bx-y=C O e.y? (exy – 1)=C of.ye «xy +bx+y=C og-yle wy+bx-y=C oh.yle xy + bx=y
Please fast back, i don't have enough time if a and are constants, the solution of Cayev+b)dx + 2ye WV + axy?owy - 1)dyo is Select one: O aye w+bx=y b.yo"4bx+y=c ce + bx-c od yox-y-c Oye "+bx-y-c oo"+bx-y=0 gye "+bx-y=c hyle-1)=0
explain please 2. Which one of the following DE is exact? a. (x+y)dx+(xy+1) dy=0 b (e + y)<x+ſe+x)dy = 0 c.(ye* +1) dx +(e' + xy) dy = 0 d. (sin x+cos y) dx +(cos x +sin y) dy = 0 e. (eº+1) dx +(e? + 2) dy = 0 3. The solution of the following separable DE xy' =-y? is a. y= '+c b. y=- c. y = In x+c In x+c d. In y=x? + e. yer+C 4....
The solution of the IVP dy dx = (ax+by+1)2 – 6; y(0)=0, where a € R and b ERVO} Select one: a. (ax +by+1)(1+x)= 1 O b. (ax+by+1)(1-x)=3 O c. (ax+by+1)2(1 - bx)=1 2 O d. (ax +by+1)= 1- bx e. (ax+by+1)(1-bx)= 1 of. (ax +by+1) (1 -bx)2 = 1 о g. (ax +by-1)(1-bx) = 1 O h. ax + by=1
Question 1 3 pts The solution of the Initial-Value Problem (IVP) S (x + y)dx – «dy = 0 is given by 1 y(1) = 0 Oy=det-1 - 1 Oy= < ln(x + y) Oy= (x + y) In x Oy= < In x None of them Question 2 3 pts The general solution of the first order non-homogeneous linear differential equation with variable coefficients dy (x + 1) + xy = e-">-1 equals dx 2 Oy=e* (C(x - 1)...