COMEDK 2023 Shift 1 Paper

To find the particular solution of the given differential equation e‌

dy

dx

=2x+1 with the initial condition y=1 when x=0, let's start by analyzing the differential equation:
The differential equation can be rewritten by taking the natural log of both sides (assuming that ‌

dy

dx

is in the range where e‌

dy

dx

is defined and 2x+1>0 ). Hence, we take the logarithm:
‌

dy

dx

=log(2x+1)
We can now integrate both sides with respect to x to find y. Integrating the right-hand side:
y=∫log(2x+1)‌dx
To solve the integral, we make the following substitution:
Let u=2x+1⇒du=2‌dx⇒dx=‌

du

2

. Therefore, substituting this in, the integral becomes:y=∫log‌u⋅‌

1

2

du=‌

1

2

‌∫log‌u‌d‌u
The integral of log‌u is a well-known result:
∫log‌u‌d‌u=u‌log‌u−u+C
Substituting back for u, we get:
Using the initial condition y(0)=1, we substitute x=0 into our equation:
1=(0+‌

1

2

)‌log(1)−0−‌

1

2

+C=−‌

1

2

+C
Now solving for C :
C=1+‌

1

2

=1.5
Therefore, the solution becomes:
y=(x+‌

1

2

)‌log(2x+1)−x−‌

1

2

+1.5=(x+‌

1

2

)‌log|2x+1|−x+1
This matches with Option A:
y=(x+‌

1

2

)‌log|2x+1|−x+1
Thus, Option A is the correct answer.