JEE Main 2022 — Mathematics Calculus

Given,

$$\frac{dy}{dx}+(\frac{2{x}^{2}+11x+13}{{x}^{3}+6{x}^{2}+11x+6})y=\frac{x+3}{x+1}$$dxdy​+(x3+6x2+11x+62x2+11x+13​)y=x+1x+3​

Now comparing with $\frac{dy}{dx}+y\times p(x)=q(x)$dxdy​+y×p(x)=q(x)

We get $p(x)=\frac{2{x}^{2}+11x+13}{{x}^{3}+6{x}^{2}+11x+6}$p(x)=x3+6x2+11x+62x2+11x+13​

So $\displaystyle IF={e}^{\int \frac{2{x}^{2}+11x+13}{{x}^{3}+6{x}^{2}+11x+6}dx}$IF=e∫x3+6x2+11x+62x2+11x+13​dx

Now finding the integration we get,

$$\int p(x)dx=\int \frac{(2{x}^{2}+11x+13)dx}{(x+1)(x+2)(x+3)}$$∫p(x)dx=∫(x+1)(x+2)(x+3)(2x2+11x+13)dx​

Using partial fraction we get,

$$\frac{2{x}^{2}+11x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}$$(x+1)(x+2)(x+3)2x2+11x+13​=x+1A​+x+2B​+x+3C​

On solving we get $A=\frac{4}{2}=2$A=24​=2, $B=1$B=1 and $C=-1$C=−1

So, $\displaystyle \int p(x)dx=A\mathrm{ln}(x+1)+B\mathrm{ln}(x+2)+c\mathrm{ln}(x+3)$∫p(x)dx=Aln(x+1)+Bln(x+2)+cln(x+3)

$$=\mathrm{ln}(\frac{{(x+1)}^{2}(x+2)}{x+3})$$=ln(x+3(x+1)2(x+2)​)

So, $\displaystyle IF={e}^{\int p(x)dx}=\frac{{(x+1)}^{2}(x+2)}{(x+3)}$IF=e∫p(x)dx=(x+3)(x+1)2(x+2)​

Now solution of differential equation is given by

$$y\times IF=\int Q\times IFdx$$y×IF=∫Q×IFdx

$$\Rightarrow y\times (\frac{{(x+1)}^{2}(x+2)}{x+3})=\int (\frac{x+3}{x+1})\frac{{(x+1)}^{2}(x+2)}{(x+3)}dx$$⇒y×(x+3(x+1)2(x+2)​)=∫(x+1x+3​)(x+3)(x+1)2(x+2)​dx

$$\Rightarrow y\times (\frac{{(x+1)}^{2}(x+2)}{x+3})=\frac{{x}^{3}}{3}+\frac{3{x}^{2}}{2}+2x+c$$⇒y×(x+3(x+1)2(x+2)​)=3x3​+23x2​+2x+c

Now given curve passes through $(0,1)$(0,1),

So, $\Rightarrow 1\times (\frac{{(0+1)}^{2}(0+2)}{0+3})=\frac{{0}^{3}}{3}+\frac{3\times {0}^{2}}{2}+2\times 0+c\Rightarrow c=\frac{2}{3}$⇒1×(0+3(0+1)2(0+2)​)=303​+23×02​+2×0+c⇒c=32​

So, curve becomes $y\times (\frac{{(x+1)}^{2}(x+2)}{x+3})=\frac{{x}^{3}}{3}+\frac{3{x}^{2}}{2}+2x+\frac{2}{3}$y×(x+3(x+1)2(x+2)​)=3x3​+23x2​+2x+32​

Now put $x=1$x=1 we get, $y\times (\frac{{(1+1)}^{2}(1+2)}{1+3})=\frac{{1}^{3}}{3}+\frac{3\times {1}^{2}}{2}+2\times 1+\frac{2}{3}$y×(1+3(1+1)2(1+2)​)=313​+23×12​+2×1+32​

$$\Rightarrow y\times 3=1+\frac{3}{2}+2\Rightarrow y\times 3=\frac{9}{2}$$⇒y×3=1+23​+2⇒y×3=29​

$$\Rightarrow y(1)=\frac{3}{2}$$⇒y(1)=23​