Let y = y(x) be the solution of the differential equation (x + 1)y' - y = e³ˣ(x + 1)², with y(0) = 1/3. Then, the point x = - 4/3 for the curve y = y(x) is :

Solution

Maths sir · 4 steps

Divide by (x+1), find the integrating factor, and the constant turns out to be zero, so y is very clean.

1. Standard linear form. Divide by x+1 to get y' - 1/(x+1) y = e³ˣ(x+1). The integrating factor is e^(-∫ dx/(x+1)) = 1/(x+1).
2. Integrate. Multiplying by the integrating factor makes the left side d/dx(y/(x+1)), and the right side becomes e³ˣ. So y/(x+1) = e³ˣ/3 + C. At x = 0, 1/3 = 1/3 + C, so C = 0.

   y = (x+1)e³ˣ/3, y' = e³ˣ(x + 4/3)

3. Check the point. y' = 0 at x = -4/3, so it is a critical point. Next, y'' = e³ˣ(3x + 5), and at x = -4/3 this gives e⁻⁴(-4 + 5) = e⁻⁴ > 0.
4. So. y' goes from negative to positive at x = -4/3, and y'' > 0 there, so it is a point of local minima.

The idea to keep. For a linear equation, put it in the form y' + Py = Q before you find the integrating factor. Then use the initial condition right away, because it often makes the constant vanish.

Answer from the NTA final answer key. Solution by PrepWiser.