Rajasthan Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितअवकल समीकरण1 Mark
Question
समीकरण $\frac{d y}{d x}=e^{x+y}+e^y x^2$ का हल है$-$
✓
Answer
$(D)\ e^x+e^{-y}+\frac{x^3}{3}=c$
$\frac{d y}{d x}=e^{x+y}+e^y x^2$
$\frac{d y}{d x}=e^x \times e^y+e^y x^2$
$\frac{d y}{d x}=e^y\left(e^x+x^2\right)$
${lr}\Rightarrow \frac{d y}{e^y}=\left(e^x+x^2\right) d x$
$\Rightarrow e^{-y} d y=\left(e^x+x^2\right) d x$
इसलिए $\int e^{-y} d y=\int\left(e^x+x^2\right) d x$
$-e^{-y}=e^x+\frac{x^3}{3}+c$
या $-e^x+e^{-y}+\frac{x^3}{3}=c$
अतः सही विकल्प $(D)$ है।
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