MCQ
જો વિકલ સમીકરણ $\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}$ નો ઉકેલ $y(0)=0$ નું સમાધાન કરે,તો $y(2)$ નું મૂલ્ય$\dots\dots\dots$છે.
- A$-1$
- B$1$
- ✓$0$
- D$e$
$=e^{e^{x}\left(x^{2}-2 x\right)}$
$\text { y. } e^{e^{x}\left(x^{2}-2 x\right)}=\int e^{e^{x}\left(x^{2}-2 x\right)} e^{x}\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{x} d x$
Let $e^{x}\left(x^{2}-2 x\right)=t$
So, $y \cdot e^{e^{c^{4}\left(x^{2}-2 x\right)}}=\int e^{t} \cdot t d t$
At $x =0, t =0$
$x =2, t =0$
$= t \cdot e ^{ t }- e ^{ t }+ c$
$x =0 ; 0 \cdot 1=0-1+ c \Rightarrow c =1$
for $x =2 ; y \cdot 1=0-1+1=0$
$y(2)=0$
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