Typical answers of 1st monthly examination

Questions Form No. ? 1 ? Solve all the ODEs
[1] y^2 (x^2-2x-8)dy-(?2y?^3+1)dx=0 [2] y^ +2y=3x
[3] (3x^2 y^3+4x)dx+(3x^3 y^2+5)dy=0 [4] y^ -?4y?^ +3y=x^2
Questions Form No. ? 2 ? Solve all the ODEs
[1] x^2 (y^2+2y-8)dx-(2x^3+1)dy=0 [2] y^ +3y=2x
[3] (3x^3 y^2+4y)dy+ (3x^2 y^3+2)dx=0 [4] y^ -3y^ +2y=x^2
Questions Form No. ? 3 ? Solve all the ODEs
[1] y^2 (x^2+2x-8)dy-(?2y?^3+3)dx=0 [2] y^ +2y=3x
[3] (3x^2 y^3+2)dx+(3x^3 y^2+2y)dy=0 [4] y^ -3y^ +2y=x^2
Questions Form No. ? 4 ? Solve all the ODEs
[1] x^2 (y^2-2y-8)dx-(2x^3+3)dy=0 [2] y^ +3y=2x
[3] (3x^3 y^2+1)dy+(3x^2 y^3+2x)dx=0 [4] y^ -?4y?^ +3y=x^2

Typical answers of 1st monthly examination
Typical answers No. ? 1 ?
[1] y^2 (x^2-2x-8)dy-(?2y?^3+1)dx=0
y^2/((?2y?^3+1) ) dy= 1/((x^2-2x-8) ) dx
1/((x^2-2x-8) )=A/((x-4) )+B/((x+2) ) ; x=4 ?A=( 1 )/( 6 ) and x=-2 ?B=-( 1 )/( 6 )
???y^2/((?2y?^3+1) ) dy?=??? (1/6(x-4) -1/6(x+2) )dx?
( 1 )/( 6 ) ln?(?2y?^3+1)=( 1 )/( 6 ) ln?(x-4)-( 1 )/( 6 ) ln?(x+2)+( 1 )/( 6 ) ln?C
?2y?^3+1=C(x-4)/( (x+2) )



[2] y^ +2y=3x
P(x)=2 and Q(x)=3x
u(x)=e^??? P(x)dx?=e^(??2 dx)=e^2x
y=1/e^2x ???3x ? e^2x dx
e^2x & I. 3x & D.
e^2x 3x

( 1 )/( 2 ) e^2x 3

( 1 )/( 4 ) e^2x 0

y=1/e^2x [( 3x )/( 2 ) e^2x-( 3 )/( 4 ) e^2x+C]=( 3x )/( 2 )-( 3 )/( 4 )+Ce^(-2x)
Another way:
y^ +2y=3x
m+2=0 ? m=-2 ? y_h=Ce^(-2x)
y_p=Ax+B ,y_p^ =A
A+2Ax+2B=3x
2A=3 ? A=( 3 )/( 2 )
A+2B=0 ? B=-( 3 )/( 4 )
y_p=( 3x )/( 2 )-( 3 )/( 4 )
y=y_h+y_p=Ce^(-2x)+( 3x )/( 2 )-( 3 )/( 4 )



[3] (3x^2 y^3+4x)dx+(3x^3 y^2+5)dy=0
?M/?y=6x^2 y^2 and ?N/?x=6x^2 y^2 ?The ODE is Exact
??(3x^2 y^3+4x)dx=x^3 y^3+2x^2+f(y)=C
??(3x^3 y^2+5)dy=x^3 y^3+5y+h(x)=C
f(y)=5y and h(x)=2x^2
So the solustion of ODE is? x^3 y^3+2x^2+5y=C
[4] y^ -?4y?^ +3y=x^2
m^2-4m+3=0
(m-1)(m-3)=0
m_1=1 ,m_2=3
y_h=c_1 e^x+c_2 e^3x
y_p=Ax^2+Bx+C then y_p^ =2Ax+B and y_p^ =2A
2A-4(2Ax+B)+3(Ax^2+Bx+C)=x^2
2A-8Ax-4B+3Ax^2+3Bx+3C=x^2
2A=1 ? A=( 1 )/( 2 )
-8A+3B=0 ? B=( 4 )/( 3 )
2A-4B+3C=0 ? C=( 13 )/( 9 )
y_p=( 1 )/( 2 ) x^2+( 4 )/( 3 ) x+( 13 )/( 9 )
y=y_h+y_p=c_1 e^x+c_2 e^3x+( 1 )/( 2 ) x^2+( 4 )/( 3 ) x+( 13 )/( 9 )






Typical answers No. ? 2 ?
[1] x^2 (y^2+2y-8)dx-(2x^3+1)dy=0
1/((y^2+2y-8) ) dy= x^2/((2x^3+1) ) dx
1/((y^2+2y-8) )=A/((y-2) )+B/((y+4) ) ; y=2 ?A=( 1 )/( 6 ) and y=-4?B=-( 1 )/( 6 )
??? (1/6(y-2) -1/6(y+4) )dy?=???x^2/((2x^3+1) ) dx?
( 1 )/( 6 ) ln?(y-2)-( 1 )/( 6 ) ln?(y+4)=( 1 )/( 6 ) ln?(2x^3+1)+( 1 )/( 6 ) ln?C
(y-2)/(y+4)=C(2x^3+1)

[2] y^ +3y=2x
P(x)=3 and Q(x)=2x
u(x)=e^??? P(x)dx?=e^(??3 dx)=e^3x
y=1/e^3x ???2x ? e^3x dx
e^3x & I. 2x & D.
e^3x 2x

( 1 )/( 3 ) e^3x 2

( 1 )/( 9 ) e^3x 0

y=1/e^3x [( 2x )/( 3 ) e^3x-( 2 )/( 9 ) e^3x+C]=( 2x )/( 3 )-( 2 )/( 9 )+Ce^(-3x)



[3] (3x^3 y^2+4y)dy+ (3x^2 y^3+2)dx=0
?M/?y=6x^2 y^2 and ?N/?x=6x^2 y^2 ?The ODE is Exact
??(3x^3 y^2+4y)dy=x^3 y^3+2y^2+f(x)=C
??(3x^2 y^3+2)dx=x^3 y^3+2x+h(y)=C
f(x)=2x and h(y)= 2y^2
So the solustion of ODE is? x^3 y^3+2y^2+2x=C

[4] y^ -3y^ +2y=x^2
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
y_p=Ax^2+Bx+C then y_p^ =2Ax+B and y_p^ =2A
2A-3(2Ax+B)+2(Ax^2+Bx+C)=x^2
2A-6Ax-3B+2Ax^2+2Bx+2C=x^2
2A=1 ? A=( 1 )/( 2 )
-6A+2B=0 ? B=( 3 )/( 2 )
2A-3B+2C=0 ? C=( 7 )/( 4 )
y_p=( 1 )/( 2 ) x^2+( 3 )/( 2 ) x+( 7 )/( 4 )
y=y_h+y_p=c_1 e^x+c_2 e^2x+( 1 )/( 2 ) x^2+( 3 )/( 2 ) x+( 7 )/( 4 )




Typical answers No. ? 3 ?
[1] y^2 (x^2+2x-8)dy-(?2y?^3+3)dx=0
y^2/((?2y?^3+3) ) dy= 1/((x^2+2x-8) ) dx
1/((x^2+2x-8) )=A/((x+4) )+B/((x-2) ) ; x=-4 ?A=-( 1 )/( 6 ) and x=2 ?B=( 1 )/( 6 )
???y^2/((?2y?^3+3) ) dy?=??? ((-1)/6(x+4) +1/6(x-2) )dx?
( 1 )/( 6 ) ln?(?2y?^3+3)=-( 1 )/( 6 ) ln?(x+4)+( 1 )/( 6 ) ln?(x-2)+( 1 )/( 6 ) ln?C
?2y?^3+3=C(x-2)/( (x+4) )
[2] y^ +2y=3x
The answer of this question is the same of Q2 in ? 1 ?
[3] (3x^2 y^3+2)dx+(3x^3 y^2+2y)dy=0
?M/?y=6x^2 y^2 and ?N/?x=6x^2 y^2 ?The ODE is Exact
??(3x^2 y^3+2)dx=x^3 y^3+2x+f(y)=C
??(3x^3 y^2+2y)dy=x^3 y^3+y^2+h(x)=C
f(y)=y^2 and h(x)=2x
So the solustion of ODE is? x^3 y^3+2x+y^2=C

[4] y^ -3y^ +2y=x^2
The answer of this question is the same of Q4 in ? 2 ?



Typical answers No. ? 4 ?
[1] x^2 (y^2-2y-8)dx-(2x^3+3)dy=0
1/((y^2-2y-8) ) dy= x^2/((2x^3+3) ) dx
1/((y^2-2y-8) )=A/((y-4) )+B/((y+2) ) ; y=4 ?A=( 1 )/( 6 ) and y=-2?B=-( 1 )/( 6 )
??? (1/6(y-4) -1/6(y+2) )dy?=???x^2/((2x^3+3) ) dx?
( 1 )/( 6 ) ln?(y-4)-( 1 )/( 6 ) ln?(y+2)=( 1 )/( 6 ) ln?(2x^3+3)+( 1 )/( 6 ) ln?C
(y-4)/(y+2)=C(2x^3+3)
[2] y^ +3y=2x
The answer of this question is the same of Q2 in ? 2 ?
[3] (3x^3 y^2+1)dy+ (3x^2 y^3+2x)dx=0
?M/?y=6x^2 y^2 and ?N/?x=6x^2 y^2 ?The ODE is Exact
??(3x^3 y^2+1)dy=x^3 y^3+y+f(x)=C
??(3x^2 y^3+2x)dx=x^3 y^3+x^2+h(y)=C
f(x)=x^2 and h(y)=2y
So the solustion of ODE is? x^3 y^3+x^2+y=C

[4] y^ -4y^ +3y=x^2
The answer of this question is the same of Q4 in ? 1 ?