A) \[y+3y+3y+y=0\]
B) \[y+3y-3y-y=0\]
C) \[y-3y-3y+y=0\]
D) \[y-3y+3y-y=0\]
Correct Answer: D
Solution :
Given,\[y=a{{e}^{x}}+bx\,\,{{e}^{x}}+c{{x}^{2}}{{e}^{x}}\] ... (i) On differentiating w.r.t. x, we get \[y=a{{e}^{x}}+b(x{{e}^{x}}+{{e}^{x}})+c({{x}^{2}}{{e}^{x}}+2x{{e}^{x}})\] \[\Rightarrow \]\[y=a{{e}^{x}}+bx{{e}^{x}}+c{{x}^{2}}{{e}^{x}}+b{{e}^{x}}+2cx{{e}^{x}}\] \[y=y+b{{e}^{x}}+2cx{{e}^{x}}\] ... (ii) Again differentiating w.r.t.\[x\], we get \[y=y+b{{e}^{x}}+2c(x{{e}^{x}}+{{e}^{x}})\] \[\Rightarrow \] \[y=y+b{{e}^{x}}+2cx{{e}^{x}}+2c{{e}^{x}}\] \[\Rightarrow \] \[y=2y-y+2c{{e}^{x}}\] ? (iii) [from Eq. (ii)] Again differentiating w.r.t.\[x\], we get \[y=2y-y+2c{{e}^{x}}\] \[\Rightarrow \] \[y=2y-y+(y=2y+y)\] [from Eq. (iii)] \[\Rightarrow \]\[y-3y+3y-y=0\]
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