WBJEE · Maths · Differential Equations
If the solution of the differential equation \(x \frac{d y}{d x}+y=x e^{x}\) be \(x y=e^{x} \phi(x)+C,\) then \(\phi(x)\) is equal to
- A \(x+1\)
- B \(x-1\)
- C \(1-x\)
- D \(x\)
Answer & Solution
Correct Answer
(B) \(x-1\)
Step-by-step Solution
Detailed explanation
Given, \(\quad x \frac{d y}{d x}+y=x e^{x}\) \(\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=e^{x}\) On comparing Eq. by \(\frac{d y}{d x}+P y=Q\), we get \(\therefore\) \[ \begin{array}{l} P=\frac{1}{x} \text { and } Q=e^{x} \\ IF=e^{\int \frac{1}{x} d x} \end{array} \]…
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