Exact Differential Equation Solved in 1 Minute | JEE Main Maths ⚡
❓ Question Let y = y ( x ) y = y(x) be the solution curve of the differential equation x ( x 2 + e x ) d y + ( e x ( x − 2 ) y − x 3 ) d x = 0 , x > 0 , x(x^2 + e^x)\,dy + \big(e^x(x - 2)y - x^3\big)\,dx = 0,\quad x>0, passing through the point ( 1 , 0 ) (1,0) . Find the value of y ( 2 ) . y(2). 🖼️ Question Image ✍️ Short Solution This is a first-order differential equation which is not exact initially . The correct JEE approach is: 👉 Convert it into a linear DE in y y 👉 Find a suitable integrating factor (IF) 👉 Use the given point to find the constant 👉 Evaluate y ( 2 ) y(2) 🔹 Step 1 — Write the DE in standard form Given: x ( x 2 + e x ) d y + ( e x ( x − 2 ) y − x 3 ) d x = 0 x(x^2 + e^x)\,dy + \big(e^x(x - 2)y - x^3\big)\,dx = 0 Rearrange: x ( x 2 + e x ) d y = ( x 3 − e x ( x − 2 ) y ) d x x(x^2 + e^x)\,dy = \big(x^3 - e^x(x - 2)y\big)\,dx Divide by x ( x 2 + e x ) x(x^2 + e^x) : d y d x + e x ( x − 2 ) x ( x 2 + e x ) y = ...