Exams Joint Entrance Examination Mathematics Differential Equations Question 439 Question 439 of 490 Medium MCQ
Let y = y ( x ) y=y(x) y = y ( x ) be the solution curve of the differential equation \[ \frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x>-1\], which passes through the point ( 0 , 1 ) (0,1) ( 0 , 1 ) . Then y ( 1 ) y(1) y ( 1 ) is equal to :
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d y d x + ( 2 x 2 + 11 x + 13 x 3 + 6 x 2 + 11 x + 6 ) y = ( x + 3 ) x + 1 , x > − 1 \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 d x d y + ( x 3 + 6 x 2 + 11 x + 6 2 x 2 + 11 x + 13 ) y = x + 1 ( x + 3 ) , x > − 1 ,
Integrating factor I.F. = e ∫ 2 x 2 + 11 x + 13 x 3 + 6 x 2 + 11 x + 6 d x =e^{\int \frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6} d x} = e ∫ x 3 + 6 x 2 + 11 x + 6 2 x 2 + 11 x + 13 d x
\[
Let 2 x 2 + 11 x + 13 ( x + 1 ) ( x + 2 ) ( x + 3 ) = A x + 1 + B x + 2 + C x + 3 A = 2 , B = 1 , C = − 1 I.F. = e ( 2 ln ∣ x + 1 ∣ + ln ∣ x + 2 ∣ − ln ∣ x + 3 ∣ ) = ( x + 1 ) 2 ( x + 2 ) x + 3 \begin{aligned}
& \text { Let } \frac{2 x^{2}+11 x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3} \\\\
& A=2, B=1, C=-1 \\\\
& \text { I.F. }=e^{(2 \ln |x+1|+\ln |x+2|-\ln |x+3|)} \\\\
& =\frac{(x+1)^{2}(x+2)}{x+3}
\end{aligned} Let ( x + 1 ) ( x + 2 ) ( x + 3 ) 2 x 2 + 11 x + 13 = x + 1 A + x + 2 B + x + 3 C A = 2 , B = 1 , C = − 1 I.F. = e ( 2 l n ∣ x + 1∣ + l n ∣ x + 2∣ − l n ∣ x + 3∣ ) = x + 3 ( x + 1 ) 2 ( x + 2 )
\]
Solution of differential equation
\[
y ⋅ ( x + 1 ) 2 ( x + 2 ) x + 3 = ∫ ( x + 1 ) ( x + 2 ) d x y ( x + 1 ) 2 ( x + 2 ) x + 3 = x 3 3 + 3 x 2 2 + 2 x + c \begin{aligned}
&y \cdot \frac{(x+1)^{2}(x+2)}{x+3}=\int(x+1)(x+2) d x \\\\
&y \frac{(x+1)^{2}(x+2)}{x+3}=\frac{x^{3}}{3}+\frac{3 x^{2}}{2}+2 x+c
\end{aligned} y ⋅ x + 3 ( x + 1 ) 2 ( x + 2 ) = ∫ ( x + 1 ) ( x + 2 ) d x y x + 3 ( x + 1 ) 2 ( x + 2 ) = 3 x 3 + 2 3 x 2 + 2 x + c
\]
Curve passes through ( 0 , 1 ) (0,1) ( 0 , 1 )
\[
\begin{gathered}
1 \times \frac{1 \times 2}{3}=0+c \Rightarrow c=\frac{2}{3} \\\\
\text { So, } y(1)=\frac{\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}}{\frac{\left(2^{2} \times 3\right)}{4}}=\frac{3}{2}
\end{gathered}
\] Answer the question to unlock