Let y=y(x) be the solution curve of the differential equation dydx+2x2+11x+13x3+6x2+11x+6y=(x+3)x+1,x>−1 which passes through the point (0,1). Then y(1) is equal to:

dydx+2x2+11x+13x3+6x2+11x+6y=x+3x+1I.F.=e∫p(x)dx∫p(x)dx=∫2x2+11x+13dx(x+1)(x+2)(x+3)Using partial fraction 2x2+11x+13(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3A=42=2B=1C=−1∵∫p(x)dx=Aln⁡(x+1)+Bln⁡(x+2)+cln⁡(x+3)=(x+1)2(x+2)x+3I.F =e∫p(x)dx=(x+1)2(x+2)(x+3)Solution y(IF)=∫Q⋅(IF)dxy(x+1)2(x+2)x+3=∫x+3x+1(x+1)2(x+2)(x+3)dx⇒y(x+1)2(x+2)x+3=x33+3x22+2x+cPasses through (0, 1) C=23Now put x=1⇒y(1)=32

Let y=y(x) be the solution curve of the differential equation dydx+2x2+11x+13x3+6x2+11x+6y=(x+3)x+1,x>−1 which passes through the point (0,1). Then y(1) is equal to:

dydx+2x2+11x+13x3+6x2+11x+6y=x+3x+1I.F.=e∫p(x)dx∫p(x)dx=∫2x2+11x+13dx(x+1)(x+2)(x+3)Using partial fraction 2x2+11x+13(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3A=42=2B=1C=−1∵∫p(x)dx=Aln⁡(x+1)+Bln⁡(x+2)+cln⁡(x+3)=(x+1)2(x+2)x+3I.F =e∫p(x)dx=(x+1)2(x+2)(x+3)Solution y(IF)=∫Q⋅(IF)dxy(x+1)2(x+2)x+3=∫x+3x+1(x+1)2(x+2)(x+3)dx⇒y(x+1)2(x+2)x+3=x33+3x22+2x+cPasses through (0, 1) C=23Now put x=1⇒y(1)=32

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

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$\frac{5}{2}$

$\frac{7}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{dy}{dx} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1} \\ I . F . = e^{\int p (x) dx} \\ \int p (x) dx = \int \frac{(2 x^{2} + 11 x + 13) dx}{(x + 1) (x + 2) (x + 3)} \end{array}$
Using partial fraction
$\begin{array}{l} \frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3} \\ A = \frac{4}{2} = 2 \\ B = 1 \\ C = - 1 \\ ∵ \int p (x) dx = Aln (x + 1) + Bln (x + 2) + cln (x + 3) \\ = (\frac{(x + 1)^{2} (x + 2)}{x + 3}) \end{array}$
I.F $= e^{\int p (x) dx} = \frac{(x + 1)^{2} (x + 2)}{(x + 3)}$
Solution $y (IF) = \int Q \cdot (IF) dx$
$\begin{array}{l} y (\frac{(x + 1)^{2} (x + 2)}{x + 3}) = \int (\frac{x + 3}{x + 1}) \frac{(x + 1)^{2} (x + 2)}{(x + 3)} dx \\ \Rightarrow y (\frac{(x + 1)^{2} (x + 2)}{x + 3}) = \frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c \end{array}$
Passes through (0, 1) $C = \frac{2}{3}$
Now put x=1
$\Rightarrow y (1) = \frac{3}{2}$