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  • Let be the solution curve of the differential equation <img alt="\mathrm{\frac{d y}{d x}+\left(\frac{2 x^{2}

Let \mathrm{ y=y(x)} be the solution curve of the differential equation
\mathrm{\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1}, which passes through the point \mathrm{(0,1).} Then \mathrm{\: y(1)} is equal to :

Option: 1

\frac{1}{2}


Option: 2

\frac{3}{2}


Option: 3

\frac{5}{2}


Option: 4

\frac{7}{2}


Answers (1)

best_answer

\begin{aligned} & \text{Using partial fractions,}\\ & \mathrm{\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}=\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}} \\ &\text { I.F. }\mathrm{=e^{2 \log (x+1)+\log (x+2)-\log (x+3)}} \\ &=\mathrm{\frac{(x+1)^{2}(x+2)}{(x+3)}}\\ &\therefore \text{Solution is}\\ &\mathrm{\frac{y(x+1)^{2}(x+2)}{(x+3)}=\int \frac{(x+3)}{(x+1)} \cdot \frac{(x+1)^{2}(x+2)}{(x+3)} d x }\\ &\mathrm{y \frac{(x+1)^{2}(x+2)}{(x+3)}=\int(x+1)(x+2) d x }\\ & \mathrm{y \frac{(x+1)^{2}(x+2)}{(x+3)}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c}\\ &\text{Using (0,1)}\\ &\mathrm{1 \cdot \frac{2}{3}=c \Rightarrow c=\frac{2}{3}}\\ &\text { Put } \mathrm{x=1 }\\ &\mathrm{y=\frac{4 \cdot 3}{4}=\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3} }\\ &\mathrm{y=\frac{1}{3}\left(\frac{27}{6}\right)=\frac{3}{2} }\\ &\therefore \operatorname{option}(B) \end{aligned}

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vishal kumar

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