JEE Mathematics 2024 Question with Solution

Given:

Find: $\alpha+2\beta+3\gamma$ if the solution is of the form

Write the differential equation as

Now take

Then

so

Using the given final form, compare with

The extracted working on the solution gives the resulting relation directly as

Applying the initial condition $y(0)=3$ is stated to determine the constants, and the final evaluated value is

Therefore, the correct option is A.

Given: $(2x + 3y - 2)dx + (4x + 6y - 7)dy = 0$ with $y(0)=3$.

Find: $\alpha+2\beta+3\gamma$.

The solution first identifies

and checks exactness:

so the equation is not exact.

It then uses the assumed form

and states that after applying the condition $y(0)=3$, the required combination evaluates to

The solution is internally incomplete in the intermediate algebra, but its final conclusion is explicit and consistent across both approaches. Hence the required value is $29$, so the correct option is A.