الاجوبة النموذجية لاسئلة الامتحان الشهري الاول 6/11/2016

(1)
1. Solve y?(x^2+1) dy=x?y dx with y(?3)=1
2. Use the test for exactness to show that the differential equation is exact, then
solve it (4x^3 y^2-6x^2 y-2x)dx+(2x^4 y-2x^3 )dy=0
3. Solve x y^ -4y=x^4+4x^2
4. Solve y^ +2y^ -15y=0 with y^ (?)=1 and y(?)=3

(2)
1.Solve x?(y^2+1) dx=y?x dy with y(1)=?3
2. Use the test for exactness to show that the differential equation is exact, then
solve it (4x^3 y^3-3x^2 y+5)dx+(3x^4 y^2-x^3-2y)dy=0
3. Solve x y^ -3y=x^3+4x
4. Solve y^ +3y^ -10y=0 with y^ (?)=1 and y(?)=4



(3)
1. Solve x?(y^2+2) dx=y?x dy with y(1)=?2
2. Use the test for exactness to show that the differential equation is exact, then
solve it (2xy^4-2y^3-3)dx+(4x^2 y^3-6xy^2+3y^2 )dy=0
3. Solve x y^ -4y=x^3+3x^2
4. Solve y^ +y^ -20y=0 with y^ (?)=-3 and y(?)=6

(4)
1. Solve y?(x^2+2) dy=x?y dx with y(?2)=1
2. Use the test for exactness to show that the differential equation is exact, then
solve it (4x^3 y^3-y^3+3x^2 )dx+(3x^4 y^2-3xy^2-2)dy=0
3. Solve x y^ -3y=x^2+2x
4. Solve y^ +3y^ -18y=0 with y^ (?)=-3 and y(?)=5

الاجوبة النموذجية (1)

1.Solve y?(x^2+1) dy=x?y dx with y(?3)=1
( y )/?y dy= x/(?(x^2+1) ) dx
? ????y dy?=1/( 2 ) ??? 2x/(?(x^2+1) )? dx
? ( 2 )/( 3 ) y^(3?2)=?(x^2+1)+C
y(?3)=1 ? ( 2 )/( 3 ) (1)^(3?2)=?((?3)^2+1)+C
? ( 2 )/( 3 )=?4+C ? C=( 2 )/( 3 )-2=-( 4 )/( 3 )
( 2 )/( 3 ) y^(3?2)=?(x^2+1)-( 4 )/( 3 ) ? 2y^(3?2)=3?(x^2+1)-4
2. Use the test for exactness to show that the differential equation is exact, then
solve it (4x^3 y^2-6x^2 y-2x)dx+(2x^4 y-2x^3 )dy=0
?M/?y=8x^3 y-6x^2 , ?N/?x=8x^3 y-6x^2 ? ?N/?x=?M/?y
Then it s exact
F=??(4x^3 y^2-6x^2 y-2x)dx=x^4 y^2-2x^3 y-x^2+w(y)=C
F=??(2x^4 y-2x^3+3)dy=x^4 y^2-2x^3 y++?(x)=C
w(y)=3y and ?(x)=-x^2
x^4 y^2-2x^3 y-x^2+3y=C







3. Solve x y^ -4y=x^4+4x^2
y^ +((-4 )/x)y=x^3+4x
P(x)=(-4 )/x and Q(x)=x^3+4x .
u(x)=e^??? P(x)dx?=e^(??(-4)?x dx)
=e^(-4 ln?x )=e^ln??x^(-4) ? =x^(-4)
y=1/u(x) ???u(x) ? Q(x)dx
y=1/x^(-4) ???x^(-4) ? (x^3+4x) dx
y=x^4 ???(( 1 )/x+( 4 )/x^3 ) ? dx
y=x^4 (ln?x-( 2 )/x^2 +C)=x^4 ln?x-2x^2+Cx^4
4. Solve y^ +2y^ -15y=0 with y^ (?)=1 and y(?)=3
m^2+2m-15=0
(m+5)(m-3)=0 ? m_1=-5 and m_2=3
y_h=c_1 e^(-5x)+c_2 e^3x
y(?)=3 ? 3=c_1 e^(-5?)+c_2 e^3? ? eq(1)
y_h^ =-5c_1 e^(-5x)+3c_2 e^3x
y^ (?)=1 ? 1=-5c_1 e^(-5?)+3c_2 e^3? ? eq(2)
eq(1)×5 ? 15=5c_1 e^(-5?)+5c_2 e^3?
Then 16=8c_2 e^3?
?(c_2=2e^(-3?) ) ? eq(3)
For eq(3) and eq(1) we get 3=c_1 e^(-5?)+2 ? ?( c_1=e^5? )
y_h=e^5? e^(-5x)+2e^(-3?) e^3x=e^5(?-x) +2e^3(x-?)




(2)
1.Solve x?(y^2+1) dx=y?x dy with y(1)=?3
x/?x dx=y/?(y^2+1) dy
????x dx?=( 1 )/( 2 ) ???2y/?(y^2+1) dy?
( 2 )/( 3 ) x^(3?2)=?(y^2+1)+C
y(1)=?3 ? ( 2 )/( 3 ) (1)^(3?2)=?((?3)^2+1)+C
( 2 )/( 3 )=?4+C ? C=( 2 )/( 3 )-2=-( 4 )/( 3 )
( 2 )/( 3 ) x^(3?2)=?(y^2+1)-( 4 )/( 3 )
2x^(3?2)=3?(y^2+1)-4
2. Use the test for exactness to show that the differential equation is exact, then
solve it (4x^3 y^3-3x^2 y+5)dx+(3x^4 y^2-x^3-2y)dy=0
?M/?y=12x^3 y^2-3x^2 , ?N/?x=12x^3 y^2-3x^2 ? ?N/?x=?M/?y
Then it s exact
F=??(4x^3 y^3-3x^2 y+5)dx=x^4 y^3-x^3 y+5x+w(y)=C
F=??(3x^4 y^2-x^3-2y)dy=x^4 y^3-x^3 y-y^2+?(x)=C
w(y)=-y^2 and ?(x)=5x
x^4 y^3-x^3 y-y^2+5x=C






3. Solve x y^ -3y=x^3+4x
y^ +((-3 )/x)y=x^2+4
P(x)=(-3 )/x and Q(x)=x^2+4 .
u(x)=e^??? P(x)dx?=e^(??(-3)?x dx)
=e^(-3 ln?x )=e^ln??x^(-3) ? =x^(-3)
y=1/u(x) ???u(x) ? Q(x)dx
y=1/x^(-3) ???x^(-3) ? (x^2+4) dx
y=x^3 ???(( 1 )/x+( 4 )/x^3 ) ? dx
y=x^3 (ln?x-( 2 )/x^2 +C)=x^3 ln?x-2x+Cx^3
4. Solve y^ +3y^ -10y=0 with y^ (?)=1 and y(?)=4
m^2+3m-10=0
(m+5)(m-2)=0 ? m_1=-5 and m_2=2
y_h=c_1 e^(-5x)+c_2 e^2x
y(?)=4 ? 4=c_1 e^(-5?)+c_2 e^2? ? eq(1)
y_h^ =-5c_1 e^(-5x)+2c_2 e^2x
y^ (?)=1 ? 1=-5c_1 e^(-5?)+2c_2 e^2? ? eq(2)
eq(1)×5 ? 20=5c_1 e^(-5?)+5c_2 e^2?
Then 21=7c_2 e^2?
?(c_2=3e^(-2?) ) ? eq(3)
For eq(3) and eq(1) we get 4=c_1 e^(-5?)+3 ? ?( c_1=e^5? )
y_h=e^5? e^(-5x)+3e^(-2?) e^2x=e^5(?-x) +3e^2(x-?)




(3)
1. Solve x?(y^2+2) dx=y?x dy with y(1)=?2
x/?x dx=y/?(y^2+2) dy
????x dx?=( 1 )/( 2 ) ???2y/?(y^2+2) dy?
( 2 )/( 3 ) x^(3?2)=?(y^2+2)+C
y(1)=?2 ? ( 2 )/( 3 ) (1)^(3?2)=?((?2)^2+2)+C
( 2 )/( 3 )=?4+C ? C=( 2 )/( 3 )-2=-( 4 )/( 3 )
( 2 )/( 3 ) x^(3?2)=?(y^2+2)-( 4 )/( 3 )
2x^(3?2)=3?(y^2+2)-4
2. Use the test for exactness to show that the differential equation is exact, then
solve it (2xy^4-2y^3-3)dx+(4x^2 y^3-6xy^2+3y^2 )dy=0
?M/?y=8xy^3-6y^2 , ?N/?x=8xy^3-6y^2 ? ?N/?x=?M/?y
Then it s exact
F=??(2xy^4-2y^3-3)dx=x^2 y^4-2xy^3-3x+w(y)=C
F=??(4x^2 y^3-6xy^2+3y^2 )dy=x^2 y^4-2xy^3+y^3+?(x)=C
w(y)=y^3 and ?(x)=-3x
x^2 y^4-2xy^3+y^3-3x=C






3. Solve x y^ -4y=x^3+3x^2
y^ +((-4 )/x)y=x^2+3x
P(x)=(-4 )/x and Q(x)=x^2+3x .
u(x)=e^??? P(x)dx?=e^(??(-4)?x dx)
=e^(-4 ln?x )=e^ln??x^(-4) ? =x^(-4)
y=1/u(x) ???u(x) ? Q(x)dx
y=1/x^(-4) ???x^(-4) ? (x^2+3x) dx
y=x^4 ???(( 1 )/x^2 +( 3 )/x^3 ) ? dx
y=x^4 ((-1)/x-( 3 )/?2x?^2 +C)=-x^3-( 3 )/( 2 ) x^2+Cx^4
4. Solve y^ +y^ -20y=0 with y^ (?)=-3 and y(?)=6
m^2+m-20=0
(m+5)(m-4)=0 ? m_1=-5 and m_2=4
y_h=c_1 e^(-5x)+c_2 e^4x
y(?)=6 ? 6=c_1 e^(-5?)+c_2 e^4? ? eq(1)
y_h^ =-5c_1 e^(-5x)+4c_2 e^4x
y^ (?)=-3 ? -3=-5c_1 e^(-5?)+4c_2 e^4? ? eq(2)
eq(1)×5 ? 30=5c_1 e^(-5?)+5c_2 e^4?
Then 27=9c_2 e^4?
?(c_2=3e^(-4?) ) ? eq(3)
For eq(3) and eq(1) we get 6=c_1 e^(-5?)+3 ? ?( c_1=3e^5? )
y_h=3e^5? e^(-5x)+3e^(-4?) e^4x=3e^5(?-x) +3e^4(x-?)




(4)
1. Solve y?(x^2+2) dy=x?y dx with y(?2)=1
( y )/?y dy= x/(?(x^2+2) ) dx
? ????y dy?=1/( 2 ) ??? 2x/(?(x^2+2) )? dx
? ( 2 )/( 3 ) y^(3?2)=?(x^2+2)+C
y(?2)=1 ? ( 2 )/( 3 ) (1)^(3?2)=?((?2)^2+2)+C
? ( 2 )/( 3 )=?4+C ? C=( 2 )/( 3 )-2=-( 4 )/( 3 )
( 2 )/( 3 ) y^(3?2)=?(x^2+2)-( 4 )/( 3 ) ? 2y^(3?2)=3?(x^2+2)-4
2. Use the test for exactness to show that the differential equation is exact, then
solve it (4x^3 y^3-y^3+3x^2 )dx+(3x^4 y^2-3xy^2-2)dy=0
?M/?y=12x^3 y^2-3y^2 , ?N/?x=12x^3 y^2-3y^2 ? ?N/?x=?M/?y
Then it s exact
F=??(4x^3 y^3-y^3+3x^2 )dx=x^4 y^3-xy^3+x^3+w(y)=C
F=??(3x^4 y^2-3xy^2-2)dy=x^4 y^3-xy^3-2y+?(x)=C
w(y)=-2y and ?(x)=x^3
x^4 y^3-xy^3+x^3-2y=C







3. Solve x y^ -3y=x^2+2x
y^ +((-3 )/x)y=x+2
P(x)=(-3 )/x and Q(x)=x+2 .
u(x)=e^??? P(x)dx?=e^(??(-3)?x dx)
=e^(-3 ln?x )=e^ln??x^(-3) ? =x^(-3)
y=1/u(x) ???u(x) ? Q(x)dx
y=1/x^(-3) ???x^(-3) ? (x+2) dx
y=x^3 ???(( 1 )/x^2 +( 2 )/x^3 ) ? dx
y=x^3 (-( 1 )/x-( 1 )/x^2 +C)=-x^2-x+Cx^3
4. Solve y^ +3y^ -18y=0 with y^ (?)=-3 and y(?)=5
m^2+3m-18=0
(m+6)(m-3)=0 ? m_1=-6 and m_2=3
y_h=c_1 e^(-6x)+c_2 e^3x
y(?)=5 ? 5=c_1 e^(-6?)+c_2 e^3? ? eq(1)
y_h^ =-6c_1 e^(-6x)+3c_2 e^3x
y^ (?)=-3 ? -3=-6c_1 e^(-6?)+3c_2 e^3? ? eq(2)
eq(1)×6 ? 30=6c_1 e^(-6?)+6c_2 e^3?
Then 27=9c_2 e^3?
?(c_2=3e^(-3?) ) ? eq(3)
For eq(3) and eq(1) we get 5=c_1 e^(-6?)+3 ? ?( c_1=2e^6? )
y_h=2e^6? e^(-6x)+3e^(-3?) e^3x=2e^6(?-x) +3e^3(x-?)