Midterm - Math 246 Show all your work for each problem (1) [18 -- 3 points each] Solve the equations: (a) dy = x 2e 4 x 4 y dx (b) 6 d2y dy 7 + 2y = 0 2 dx dx (c) dy = 3x 2 (1+ y 2 ) dx (d) (y 2 3x 2 )dy + (2xy)dx = 0 (e) (x 2 + y 2 )dx + (2xy)dy = 0 (f) dy y + = 5(x 2) y dx x 2 (2) [12 -- 3 points each] Solve the initial value problems: (a) dy = 8x 3e2 y ; y(1) = 0 dx (b) d2y dy + 2 + 10y = 0 ; y(0) = 1, y'(0) = 1 2 dx dx (c) y''+ 6y'+ 9y = 0 ; y(0) = 2, y'(0) = (d) 25 3 dy y + + 2 = 3x ; y(1) = 1 dx x 1 (3) [8] Newton's Law of Cooling assumes that the rate of cooling is proportional to the temperature difference (M-T) between the medium (environment) temperature M and the object's temperature T, dT = k(M T) where k is the constant of proportionality. Do the which leads to the equation: dt following: (a) Solve this equation for T. (b) If a pot of boiling water (100 degrees Celsius) is removed from the heat and allowed to cool. After five minutes the temperature is 80 degree C, and in another five minutes it has dropped to 65 deg. What is the temperature of the kitchen? (4) [8] Given the initial value problem: dy = (x y) ; y(0) = 0 dx Do the following: (a) Verify that y = e x + x 1 is the (why \"the\") solution for this problem. (b) Using the Improved Euler Method (with h=0.1), approximate the solution on the interval [0,1]. (c) Graph the approximation with the actual solution for comparison. (5) [4] Determine the effects of changing the parameter b in the initial value problem: y''+ by'+ 9y = 0 ; y(0) = 1, y'(0) = 0 Graph the solutions for b=1,2,3,4,5 for comparison. 2 Surname 1 Name Professor Course Date Mathematical Computations Question 1 dy 2 4 x =x e 4 y dx a. dy 2 4 x =x e 4 y dx x dy=( 2 e4 x 4 y ) dx x ( 2 e 4 y )dx dy= 4 x x 3 e4 x 2 2 x y= x e 4 y x +c 3 2 b. d2 y dy 6 2 7 +2 y=0 dx dx 6 d2 y dy 7 +2 y=0 2 dx dx dy dy 6 7 =2 y dx dx ( ( y 6 ) dy 7 =2 yx dx ) (6 dydx 7)=2 x Surname 2 6 dy =2 x+ 7 dx 6 dy= (2 x+7 ) dx 2 x 6 y=2 +7 x 2 2 x 7 y= + x+ c 6 6 c. dy =3 x 2 (1+ y 2) dx dy=( 3 x 2 (1+ y 2) ) dx dy = (3 x 2 (1+ y 2 )) dx y=x 3 ( 1+ y 2 ) 3 x 2 ( x+ xy 2) y=x 3x 3 y 23 x 33 x 3 y 2 3 3 2 y=4 x 4 x y + c 3 2 y=4 x (1+ y )+c d. ( y 2 3 x 2 ) dy+ (2 xy ) dx=0 ( y 2 3 x 2 ) dy+ (2 xy ) dx=0 dy ( 2 xy ) = dx ( y 2 3 x 2 ) dy= (( ( 2 xy ) y 23 x 2) dy = y= (( ) dx ( 2 xy ) y 2 3 x 2 ) x2 1 +c y3 y ) dx Surname 3 e. ( x 2 + y 2 ) d x + ( 2 xy ) d y=0 ( x 2 + y 2 ) dy= dx 2 xy dy = ( ( x2 + y 2 ) dx 2 xy ) y=( x 2 + y 2 ) ln ( xy ) ( x2 y ) f. x2 y +c 6y 2 dy y + =5(x2) y dx x2 dy y + =5(x2) y dx x2 dy y =5 ( x2 ) y dx x2 ( dy= 5 ( x2 ) y y dx x2 ) y dy = (5 ( x2 ) y x2 ) dx y= ( 52 ( x2) y x (x 2 x ) yln ( x 2) xy )+c 2 Question 2 a. dy =8 x 3 e2 y ; y ( 1 )=0 dx dy =8 x3 e 2 y dx dy=( 8 x 3 e 2 y ) dx dy = ( 8 x 3 e 2 y ) dx Surname 4 y=8 x 3 e2 yx +2 x 4 e 2 y + c When x = 1, y = 0. 0=10+c c=10 3 2 yx y=8 x e 4 2y +2 x e 10 y=4 x 3 e 2 yx + x 4 e 2 y 5 b. d2 y dy + 2 + 10 y =0 ; y ( 0 )=1, y ' ( 0 ) 1 2 dx dx dy dy +2 =10 y dx dx ( dy y ) ( dydx + 2)=10 ydx ( dydx +2)=10 yx +c When x = 0, y = 1. ( dydx +2)=c y ( dydx +2)=10 yx +( dydx +2) y dy dy +2 y=10 yx + +2 dx dx y dy dy =10 xy 2 y+2 dx dx ( dy ( y1 )=10 xy2 y +2 dx ) Surname 5 dy 10 xy2 y+ 2 = dx y 1 dy= 10 xy 2 y+2 dx y1 2 1 5 x y4 x y+ 4 x y= +c y1 When x = 0, y = -1. 1= 00+ 0 +c 11 1=c y= 15 x 2 y4 xy+ 4 x 1 y1 c. y''6y'9y 0 ; y(0) 2, y'(0) 25/3 '' ' y +6 y +9 y=0 d2 y dy +6 +9 y=0 2 dx dx dy dy +6 =9 y dx dx ( dy y ) ( dydx +6)=9 ydx ( dydx + 6)=9 yx+ c When x = 0, y = 2. 2 ( dydx + 6)=c Surname 6 y ( dydx + 6)=9 yx+ 2( dydx +6) dy 9 yx = 6 dx y +2 dy= y= 9 yx 6 dx y +2 9 yx 6 x +c 2 ( y+2 ) When x = 0, y = 25/3. 25 =c 3 y= d. 9 yx 25 6 x + 3 2 ( y+2 ) dy y + +2=3 x ; y ( 1 )=1 dx x dy y + +2=3 x dx x dy y =3 x 2 dx x y dy= 3 x 2 dx x ( ) 3 y= x 2 y ex 2 x+ c 2 When x = 1, y = 1. 3 1= e12+c 2 c=2.13 2 x y=1.5 x y e 2 x2.13 Surname 7 Question 3 a. Solve this equation for T dT =k ( M T ) dt dT =kdt ( M T ) dln(M T )=kdt ln ( M T )=K +kt We know that M=T at t=0. Hence ln (M 0 T )=K Therefore ln ( M T )=ln ( M 0T ) kt ln ( M T ) =kt ( M 0T ) ( M T ) =ekt ( M 0 T ) ( M T )=( M 0T ) ekt T =M ( M 0T ) ekt b. Numerical integration T =M ( M 0T ) ekt dT ( M iT i+1 ) dt t ( M i T i +1) t k (M iT i ) T i +1=M ik ( M i T i ) t 80=175 M i175 M=1.45 Surname 8 Question 4 Initial value problem dy =( x y ) ; y ( 0 ) =0 dx x a. Verify that y=e + x1 dy =( x y ) dx dy=( x y ) dx dy = ( x y ) dx 1 y= x 2 yx +c 2 When x = 0, y = 0. Hence, x=y 1=11+ c c=1 x y=e + x1 b. Improved Euler method X1 = x0 + h X1=0+0.1=0.1 y=0.9 Initial value y x Question 5 y''by'9y 0 ; y(0) 1, y'(0) 0 Surname 9 d2 y dy + b +9 y=0 2 dx dx dy dy +b =9 y dx dx ( dy y ) ( dydx + b)=9 ydx ( dydx + b)=9 yx+ c When x = 0, y = 1. ( dydx + b)=c y ( dydx + b)=9 yx+( dydx + b) dy 9 yx = 6 dx y +2 dy= y= 9 yx b dx y +1 9 yx b x +c 2 ( y+1 ) When x = 0, y = 0. 0=c y= 9 yx b x 2 ( y+2 ) b=1,2,3,4,5 Surname 10 y x