JEE Main 2023 — Mathematics Calculus

Given differential equation is $\frac{dy}{dx}+\frac{4x}{({x}^{2}-1)}y=\frac{x+2}{{({x}^{2}-1)}^{\frac{5}{2}}},x>1$dxdy​+(x2−1)4x​y=(x2−1)25​x+2​,x>1

Now $\displaystyle \mathrm{IF}={e}^{\int \frac{4x}{{x}^{2}-1}dx}={({x}^{2}-1)}^{2}$IF=e∫x2−14x​dx=(x2−1)2

The required equation will be $\displaystyle \Rightarrow y\cdot {({x}^{2}-1)}^{2}=\int \frac{x+2}{{({x}^{2}-1)}^{1/2}}dx$⇒y⋅(x2−1)2=∫(x2−1)1/2x+2​dx

$$\Rightarrow y\cdot {({x}^{2}-1)}^{2}=\frac{1}{2}\int \frac{2x}{{({x}^{2}-1)}^{1/2}}dx+2\int \frac{dx}{{({x}^{2}-1)}^{1/2}}$$⇒y⋅(x2−1)2=21​∫(x2−1)1/22x​dx+2∫(x2−1)1/2dx​

$$\Rightarrow y\cdot {({x}^{2}-1)}^{2}=2\mathrm{ln}(|\sqrt{{x}^{2}-1}+x|)+\sqrt{{x}^{2}-1}+C$$⇒y⋅(x2−1)2=2ln(∣x2−1​+x∣)+x2−1​+C

Now using, at $x=2,$x=2, $y(2)=\frac{2}{9}{\mathrm{log}}_{e}(2+\sqrt{3})$y(2)=92​loge​(2+3​) we get,

$$\Rightarrow 9\cdot \frac{2}{9}\mathrm{ln}(2+\sqrt{3})=2\mathrm{ln}(2+\sqrt{3})+\sqrt{3}+C$$⇒9⋅92​ln(2+3​)=2ln(2+3​)+3​+C

$$\Rightarrow C=-\sqrt{3}$$⇒C=−3​

Now finding the value of function at $x=\sqrt{2}$x=2​ we get,

$$y\times 1=2\mathrm{ln}(1+\sqrt{2})+1-\sqrt{3}$$y×1=2ln(1+2​)+1−3​

Now on comparing we get,

$$\Rightarrow \beta =1,\alpha =2,\gamma =3$$⇒β=1,α=2,γ=3

$$\Rightarrow \alpha \beta \gamma =1\times 2\times 3=6$$⇒αβγ=1×2×3=6

Hence, this is the required answer.