Question

Solve the given initial value problem and determine at least approximately where the solution is valid.

(12x2+y−1)dx−(18y−x)dy=0, y(1)=0

Chapter 2, Section 2.6, Question 10 Solve the given initial value problem and determine at least approximately where the solution is valid. (12x2 + y − 1) dx – (18y – x) d y = 0, y(1) = 0 y = the solution is valid as long as Q@20

Answer 1

Sol. (1930?+ 4-1)dx- (184 - x) dx = 0 , 401): 0 Here M(1,4)= 19x +y-1, N(2,4)= -( 184. ) 2 Jon = 1 and JN - 2 so given diff. ean. exact and General Solution F(2,4)= (mda +( ndy where N is function of y only F(1,9) = [1232? +9-1]dx-fisy dy = 423 + Bey--9y2 General solution of equation 4x + x - 2 -quºc c. As given 4(1)=0 = 4 +0-1-0=C -9y?+ Dey - x +48-3 - 0 ou qy? - 24+(x-423+3) = 0 Y= x + 1-)-3662-48°43) 04 1:3) 18 on y: 162-362 +144843-108 > y = x + 514433+*2-362-108 18

but Jy x-5144x3+x2-362-108 solution be cause it satisfy initial condition 4(1)=0 - Now , tue ie. number in Discriminant square root must be solution exist when 144313 +2-362-108 20