JEE Mains · Maths · STD 12 - 9. differential equations
If \(y(x)\) is the solution of the differential equation \((x + 2)\frac{{dy}}{{dx}} = {x^2} + 4x - 9,\,x \ne - 2\) and \(y(0) = 0,\) then \(y(-4)\) is equal to
- A \(0\)
- B \(2\)
- C \(1\)
- D \(-1\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\(\left( {x + 2} \right)\frac{{dy}}{{dx}} = {x^2} + 4x - 9x \ne - 2\) \(\frac{{dy}}{{dx}} = \frac{{{x^2} + 4x - 9}}{{x + 2}}\) \(\mathrm{dy}=\begin{array}{c}{x^{2}+4 x-9} \\ {x+2}\end{array} \mathrm{d} x\) \(\int d y=\int \frac{x^{2}+4 x-9}{x+2} d x\)…
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