MHT CET 2021 Math Solved Paper

Given, differential equation
‌

dy

dx

=‌

y+1

x2−x

⇒‌

dy

y+1

=‌

dx

x2−x

On integrating both sides, we get
∫‌

dy

y+1

=∫‌

dx

x2−x

Now, ‌

1

x2−x

=‌

1

x(x−1)

=‌

A

x

+‌

B

x−1

∴‌‌‌

1

x(x−1)

‌‌=‌

A

x

+‌

B

x−1

. . . (i)
⇒‌‌1‌‌=A(x−1)+B(x)
Putting x=0, then
‌1=A(0−1)⇒A=−1
‌ Putting ‌x−1=0,‌‌ then ‌x=1
∴‌1=A(0)+B(1)
⇒‌B=1
From Eq. (i), we get
‌‌∫‌

dy

y+1

=∫−‌

1dx

x

+∫‌

1

x−1

dx
⇒‌‌log(y+1)=−log‌x+log(x−1)+log‌C
⇒log(y+1)+log‌x−log(x−1)=log‌C
⇒‌‌log{‌

x(y+1)

x−1

}=log‌C
⇒‌‌‌

x(y+1)

x−1

=C . . . (ii)
On putting x=2 and y=1 in Eq. (ii), we get
‌

2(1+1)

2−1

=C⇒C=(2)(2)=4
Putting value of C=4 in Eq. (ii), we get
‌

x(y+1)

x−1

=4
⇒‌‌xy+x=4x−4‌‌⇒‌‌xy=3x−4