Typical answers for 1st monthly examination of 2nd class

Questions Form No. ? 1 ? Solve all the ODEs
[1] y(x^2-2x-15)dy-(?4y?^2+1)dx=0 [2] (xy^2+4)dx+(x^2 y+3)dy=0
[3] x^2 y^ -xy=2 [4a] y^ -?5y?^ +6y=0
[4b] y^ -6y^ +13y=0 [4c] y^ +6y^ +9y=0

Questions Form No. ? 2 ? Solve all the ODEs
[1] x(y^2+2y-15)dx-(4x^2+1)dy=0 [2] (x^2 y+4)dy+(xy^2+3)dx=0
[3] x^2 y^ -xy=3 [4a] y^ -?5y?^ +4y=0
[4b] y^ -6y^ +9y=0 [4c] y^ +6y^ +13y=0

Questions Form No. ? 3 ? Solve all the ODEs
[1] 2y(x^2+x-6)dy-(?5y?^2+3)dx=0 [2] (xy^2+3)dx+(x^2 y+2)dy=0
[3] x^2 y^ -xy=1 [4a] y^ -?6y?^ +9y=0
[4b] y^ -3y^ +2y=0 [4c] y^ +6y^ +13y=0

Questions Form No. ? 4 ? Solve all the ODEs
[1] 2x(y^2-y-6)dx-(5x^2+4)dy=0 [2] (x^2 y+2)dy+(xy^2+1)dx=0
[3] x^2 y^ -xy=4 [4a] y^ +?6y?^ +9y=0
[4b] y^ -6y^ +13y=0 [4c] y^ -4y^ +3y=0







Typical answers No. ? 1 ?
[1] y(x^2-2x-15)dy-(4y^2+1)dx=0
y/((4y^2+1) ) dy= 1/((x^2-2x-15) ) dx
1/((x^2-2x-15) )=A/((x-5) )+B/((x+3) ) ; x=5 ?A=( 1 )/( 8 ) and x=-3 ?B=-( 1 )/( 8 )
???y/((4y^2+1) ) dy?=??? (1/8(x-5) -1/8(x+3) )dx?
( 1 )/( 8 ) ln?(4y^2+1)=( 1 )/( 8 ) ln?(x-5)-( 1 )/( 8 ) ln?(x+3)+( 1 )/( 8 ) ln?C
4y^2+1=C(x-5)/( (x+3) )
[2] (xy^2+4)dx+(x^2 y+3)dy=0
?M/?y=2xy and ?N/?x=2xy ?The ODE is Exact
??(xy^2+4)dx=(x^2 y^2)/2+4x+f(y)=C
??(x^2 y+3)dy=(x^2 y^2)/2+3y+h(x)=C
f(y)=3y and h(x)=4x
So the solustion of ODE is? (x^2 y^2)/2+3y+4x=C
[3] x^2 y^ -xy=2
y^ -( 1 )/( x ) y=2/x^2 ; P(x)=-( 1 )/x and Q(x)=2/x^2
u(x)=e^??? P(x)dx?=e^(??(-1)?x dx)=e^(-ln?x )=e^ln??x^(-1) ? =x^(-1)=( 1 )/x
y=1/(1?x) ???( 1 )/x ?×2/x^2 dx =x???2/x^3 dx ?=x[-1/x^2 +C]
y=Cx-( 1 )/x

[4a] y^ -?5y?^ +6y=0
m^2-5m+6=0
(m-3)(m-2)=0
m_1=3 ,m_2=2
y_h=c_1 e^3x+c_2 e^2x
[4b] y^ -6y^ +13y=0
m^2-6m+13=0
m=(-b??(b^2-4ac))/2a=(-(-6)??((-6)^2-4×1×13))/(2×1)
=(6??(36-52))/2=(6??(-16))/2=3?2i
y_h=e^3x (c_1 sin?2x+c_2 cos?2x )
[4c] y^ +6y^ +9y=0
m^2+6m+9=0
(m+3)(m+3)=0
m_1=m_2=-3
y_h=(c_1 x+c_2 ) e^(-3x)
















Typical answers No. ? 2 ?
[1] x(y^2+2y-15)dx-(4x^2+1)dy=0
1/((y^2+2y-15) ) dy= x/((4x^2+1) ) dx
1/((y^2+2y-15) )=A/((y-3) )+B/((y+5) ) ; y=3 ?A=( 1 )/( 8 ) and y=-5 ?B=-( 1 )/( 8 )
??? (1/8(y-3) -1/8(y+5) )dy?=???x/((4x^2+1) ) dx?
( 1 )/( 8 ) ln?(y-3)-( 1 )/( 8 ) ln?(y+5)=( 1 )/( 8 ) ln?(4x^2+1)+( 1 )/( 8 ) ln?C
(y-3)/(y+5)=C(4x^2+1)
[2] (x^2 y+4)dy+(xy^2+3)dx=0
?M/?y=2xy and ?N/?x=2xy ?The ODE is Exact
??(xy^2+4)dy=(x^2 y^2)/2+4y+f(x)=C
??(x^2 y+3)dx=(x^2 y^2)/2+3x+h(y)=C
f(x)=3x and h(y)=4y
So the solustion of ODE is? (x^2 y^2)/2+3x+4y=C
[3] x^2 y^ -xy=3
y^ -( 1 )/( x ) y=3/x^2 ; P(x)=-( 1 )/x and Q(x)=3/x^2
u(x)=e^??? P(x)dx?=e^(??(-1)?x dx)=e^(-ln?x )=e^ln??x^(-1) ? =x^(-1)=( 1 )/x
y=1/(1?x) ???( 1 )/x ?×3/x^2 dx =x???3/x^3 dx ?=x[-3/?2x?^2 +C]
y=Cx-( 3 )/2x
[4a] y^ -?5y?^ +4y=0
m^2-5m+4=0
(m-4)(m-1)=0
m_1=4 ,m_2=1
y_h=c_1 e^4x+c_2 e^x
[4b] y^ -6y^ +9y=0
m^2-6m+9=0
(m-3)(m-3)=0
m_1=m_2=3
y_h=(c_1 x+c_2 ) e^3x
[4c] y^ +6y^ +13y=0
m^2+6m+13=0
m=(-b??(b^2-4ac))/2a=(-6??((6)^2-4×1×13))/(2×1)=-3?2i
y_h=e^(-3x) (c_1 sin?2x+c_2 cos?2x )












Typical answers No. ? 3 ?
[1] 2y(x^2+x-6)dy-(5y^2+3)dx=0
2y/((5y^2+3) ) dy= 1/((x^2+x-6) ) dx
1/((x^2+x-6) )=A/((x-2) )+B/((x+3) ) ; x=2 ?A=( 1 )/( 5 ) and x=-3 ?B=-( 1 )/( 5 )
???2y/((5y^2+3) ) dy?=??? (1/5(x-2) -1/5(x+3) )dx?
( 1 )/( 5 ) ln?(5y^2+3)=( 1 )/( 5 ) ln?(x-2)-( 1 )/( 5 ) ln?(x+3)+( 1 )/( 5 ) ln?C
5y^2+3=C(x-2)/( (x+3) )
[2] (xy^2+3)dx+(x^2 y+2)dy=0
?M/?y=2xy and ?N/?x=2xy ?The ODE is Exact
??(xy^2+3)dx=(x^2 y^2)/2+3x+f(y)=C
??(x^2 y+2)dx=(x^2 y^2)/2+2y+h(x)=C
h(x)=3x and f(y)=2y
So the solustion of ODE is? (x^2 y^2)/2+3x+2y=C
[3] x^2 y^ -xy=1
y^ -( 1 )/( x ) y=1/x^2 ; P(x)=-( 1 )/x and Q(x)=1/x^2
u(x)=e^??? P(x)dx?=e^(??(-1)?x dx)=e^(-ln?x )=e^ln??x^(-1) ? =x^(-1)=( 1 )/x
y=1/(1?x) ???( 1 )/x ?×1/x^2 dx =x???1/x^3 dx ?=x[-1/?2x?^2 +C]
y=Cx-( 1 )/2x

[4a] y^ -?6y?^ +9y=0
m^2-6m+9=0
(m-3)(m-3)=0
m_1=m_2=3
y_h=(c_1 x+c_2 ) e^3x
[4b] y^ -3y^ +2y=0
m^2-3m+2=0
(m-2)(m-1)=0
m_1=2 ,m_2=1
y_h=c_1 e^2x+c_2 e^x
[4c] y^ +6y^ +13y=0
m^2+6m+13=0
m=(-b??(b^2-4ac))/2a=(-6??((6)^2-4×1×13))/(2×1)=-3?2i
y_h=e^(-3x) (c_1 sin?2x+c_2 cos?2x )











Typical answers No. ? 4 ?
[1] 2x(y^2-y-6)dx-(5x^2+4)dy=0
1/((y^2-y-6) ) dy= 2x/((5x^2+4) ) dx
1/((y^2-y-6) )=A/((y-3) )+B/((y+2) ) ; y=3 ?A=( 1 )/( 5 ) and y=-2 ?B=-( 1 )/( 5 )
??? (1/5(y-3) -1/5(y+2) )dy?=???2x/((5x^2+4) ) dx?
( 1 )/( 5 ) ln?(y-3)-( 1 )/( 5 ) ln?(y+2)=( 1 )/( 5 ) ln?(5x^2+4)+( 1 )/( 5 ) ln?C
(y-3)/(y+2)=C(5x^2+4)

[2] (x^2 y+2)dy+(xy^2+1)dx=0
?M/?y=2xy and ?N/?x=2xy ?The ODE is Exact
??(xy^2+2)dy=(x^2 y^2)/2+2y+f(x)=C
??(x^2 y+1)dx=(x^2 y^2)/2+x+h(y)=C
f(x)=x and h(y)=2y
So the solustion of ODE is? (x^2 y^2)/2+x+2y=C
[3] x^2 y^ -xy=4
y^ -( 1 )/( x ) y=4/x^2 ; P(x)=-( 1 )/x and Q(x)=4/x^2
u(x)=e^??? P(x)dx?=e^(??(-1)?x dx)=e^(-ln?x )=e^ln??x^(-1) ? =x^(-1)=( 1 )/x
y=1/(1?x) ???( 1 )/x ?×4/x^2 dx =x???4/x^3 dx ?=x[-2/x^2 +C]
y=Cx-( 2 )/x

[4a] y^ +?6y?^ +9y=0
m^2+6m+9=0
(m+3)(m+3)=0
m_1=m_2=-3
y_h=(c_1 x+c_2 ) e^(-3x)
[4b] y^ -6y^ +13y=0
m^2-6m+13=0
m=(-b??(b^2-4ac))/2a=(-(-6)??((-6)^2-4×1×13))/(2×1)=3?2i
y_h=e^3x (c_1 sin?2x+c_2 cos?2x )
[4c] y^ -4y^ +3y=0
m^2-4m+3=0
(m-3)(m-1)=0
m_1=3 ,m_2=1
y_h=c_1 e^3x+c_2 e^x