Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Find the particular solution of $\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1,$ that $\text{y}=3,$ when $\text{x}=0.$
✓
Answer
$\text{e}^{\frac{\text{dy}}{\text{dx}}}=\text{x}+1$ $\frac{\text{dy}}{\text{dx}}=\log(\text{x}+1),\text{y}=3$ at $\text{x}=0$ $\int\text{dy}=\int\log(\text{x}+1)\text{dx}$ $\text{y}=\log|\text{x}+1|\times\int1\times\text{dx}-\int\Big(\frac{1}{\text{x}+1}\times\int1\text{dx}\Big)\text{dx}+\text{C}$ Using integration by parts $\text{y = x}\log|\text{x}+1|-\int\frac{\text{x}}{\text{x}+1}\text{dx}+\text{C}$ $\text{y = x}\log|\text{x}+1|-\Big(\int\Big(1-\frac{1}{\text{x}+1}\Big)\text{dx}\Big)+\text{C}$ $=\text{x}\log|\text{x}+1|-(\text{x}-\log|\text{x}+1|)+\text{C}$ $\text{y = x}\log|\text{x}+1|-\text{x}+\log|\text{x}+1|+\text{C}$ $\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x + C}$ Put $\text{y}=3$ and $\text{x}=0$ $3=0-0+\text{C}$ $\text{C}=3$ Put $\text{C}=3$ in equation (1), $\text{y}=(\text{x}+1)\log|\text{x}+1|-\text{x}+3$
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