If 2x = y ^1 \over 5 + y ^ - 1 \over 5 and (x2 - 1) d^2y \over dx^2 + \lambda x dy \over dx + ky = 0, then \lambda + k is equal to :

Differential Equations PYQ — JEE Main 2017 · 9 Apr · Shift 1 · Q38 with Solution

View written solutionFree

Correct answer: B

Given relation

We have $2x = y^{1/5} + y^{-1/5}.$

Let $t = y^{1/5}.$ Then

\qquad 2x = t + \frac{1}{t}.$$ So we will express everything in terms of $t$. --- 2. **Differentiate $x$ with respect to $t$ and $y$ with respect to $t$** From $$x = \frac{1}{2}\left(t + \frac{1}{t}\right),$$ we get $$\frac{dx}{dt} = \frac{1}{2}\left(1 - \frac{1}{t^2}\right)=\frac{t^2-1}{2t^2}.$$ Also, $$y=t^5 \implies \frac{dy}{dt}=5t^4.$$ Hence $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{5t^4}{(t^2-1)/(2t^2)} = \frac{10t^6}{t^2-1}.$$ --- 3. **Find $\dfrac{d^2y}{dx^2}$** Using $$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \Big/ \frac{dx}{dt},$$ first differentiate $$\frac{dy}{dx}=\frac{10t^6}{t^2-1}.$$ Now, $$\frac{d}{dt}\left(\frac{10t^6}{t^2-1}\right) =10\cdot \frac{6t^5(t^2-1)-t^6(2t)}{(t^2-1)^2}.$$ Simplify the numerator: $$6t^5(t^2-1)-2t^7 = 6t^7-6t^5-2t^7 = 4t^7-6t^5 = 2t^5(2t^2-3).$$ So, $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{20t^5(2t^2-3)}{(t^2-1)^2}.$$ Therefore, $$\frac{d^2y}{dx^2} = \frac{20t^5(2t^2-3)}{(t^2-1)^2} \cdot \frac{2t^2}{t^2-1} =\frac{40t^7(2t^2-3)}{(t^2-1)^3}.$$ --- 4. **Express $x^2-1$ in terms of $t$** Since $$2x=t+\frac1t,$$ we have $$4x^2 = t^2+2+\frac1{t^2}.$$ So, $$4(x^2-1)=t^2-2+\frac1{t^2}=\left(t-\frac1t\right)^2.$$ Also, $$t+\frac1t=2x \implies \left(t-\frac1t\right)^2=4(x^2-1).$$ But directly, $$x^2-1=\frac{1}{4}\left(t^2-2+\frac1{t^2}\right) =\frac{(t^2-1)^2}{4t^2}.$$ --- 5. **Compute $(x^2-1)\dfrac{d^2y}{dx^2}$** Using $$x^2-1=\frac{(t^2-1)^2}{4t^2},$$ we get $$(x^2-1)\frac{d^2y}{dx^2} = \frac{(t^2-1)^2}{4t^2}\cdot \frac{40t^7(2t^2-3)}{(t^2-1)^3} = \frac{10t^5(2t^2-3)}{t^2-1}.$$ Also, $$x\frac{dy}{dx} = \frac12\left(t+\frac1t\right)\cdot \frac{10t^6}{t^2-1} =\frac{5t^6\left(t+\frac1t\right)}{t^2-1} =\frac{5t^5(t^2+1)}{t^2-1}.$$ And $$y=t^5.$$ --- 6. **Substitute into the differential equation** Given $$(x^2-1)\frac{d^2y}{dx^2}+\lambda x\frac{dy}{dx}+ky=0.$$ Substitute the expressions: $$\frac{10t^5(2t^2-3)}{t^2-1}+\lambda\cdot \frac{5t^5(t^2+1)}{t^2-1}+kt^5=0.$$ Factor out $t^5$: $$t^5\left[\frac{10(2t^2-3)+5\lambda(t^2+1)}{t^2-1}+k\right]=0.$$ Multiply by $(t^2-1)$: $$10(2t^2-3)+5\lambda(t^2+1)+k(t^2-1)=0.$$ Expand: $$20t^2-30+5\lambda t^2+5\lambda+kt^2-k=0.$$ Group coefficients: $$(20+5\lambda+k)t^2+(-30+5\lambda-k)=0.$$ Since this holds for all $t$, both coefficients must be zero: $$20+5\lambda+k=0 \quad ...(1)$$ $$-30+5\lambda-k=0 \quad ...(2)$$ Add (1) and (2): $$10\lambda-10=0 \implies \lambda=1.$$ Then from (1): $$20+5(1)+k=0 \implies 25+k=0 \implies k=-25.$$ Thus, $$\lambda+k=1+(-25)=-24.$$ --- 7. **Check options** The value is $$-24.$$ So the correct option is: **B: $-24$** --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer is also **B**. Hence they agree.

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View written solutionFree

Correct answer: B

Given relation

We have $2x = y^{1/5} + y^{-1/5}.$

Let $t = y^{1/5}.$ Then

\qquad 2x = t + \frac{1}{t}.$$ So we will express everything in terms of $t$. --- 2. **Differentiate $x$ with respect to $t$ and $y$ with respect to $t$** From $$x = \frac{1}{2}\left(t + \frac{1}{t}\right),$$ we get $$\frac{dx}{dt} = \frac{1}{2}\left(1 - \frac{1}{t^2}\right)=\frac{t^2-1}{2t^2}.$$ Also, $$y=t^5 \implies \frac{dy}{dt}=5t^4.$$ Hence $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{5t^4}{(t^2-1)/(2t^2)} = \frac{10t^6}{t^2-1}.$$ --- 3. **Find $\dfrac{d^2y}{dx^2}$** Using $$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \Big/ \frac{dx}{dt},$$ first differentiate $$\frac{dy}{dx}=\frac{10t^6}{t^2-1}.$$ Now, $$\frac{d}{dt}\left(\frac{10t^6}{t^2-1}\right) =10\cdot \frac{6t^5(t^2-1)-t^6(2t)}{(t^2-1)^2}.$$ Simplify the numerator: $$6t^5(t^2-1)-2t^7 = 6t^7-6t^5-2t^7 = 4t^7-6t^5 = 2t^5(2t^2-3).$$ So, $$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{20t^5(2t^2-3)}{(t^2-1)^2}.$$ Therefore, $$\frac{d^2y}{dx^2} = \frac{20t^5(2t^2-3)}{(t^2-1)^2} \cdot \frac{2t^2}{t^2-1} =\frac{40t^7(2t^2-3)}{(t^2-1)^3}.$$ --- 4. **Express $x^2-1$ in terms of $t$** Since $$2x=t+\frac1t,$$ we have $$4x^2 = t^2+2+\frac1{t^2}.$$ So, $$4(x^2-1)=t^2-2+\frac1{t^2}=\left(t-\frac1t\right)^2.$$ Also, $$t+\frac1t=2x \implies \left(t-\frac1t\right)^2=4(x^2-1).$$ But directly, $$x^2-1=\frac{1}{4}\left(t^2-2+\frac1{t^2}\right) =\frac{(t^2-1)^2}{4t^2}.$$ --- 5. **Compute $(x^2-1)\dfrac{d^2y}{dx^2}$** Using $$x^2-1=\frac{(t^2-1)^2}{4t^2},$$ we get $$(x^2-1)\frac{d^2y}{dx^2} = \frac{(t^2-1)^2}{4t^2}\cdot \frac{40t^7(2t^2-3)}{(t^2-1)^3} = \frac{10t^5(2t^2-3)}{t^2-1}.$$ Also, $$x\frac{dy}{dx} = \frac12\left(t+\frac1t\right)\cdot \frac{10t^6}{t^2-1} =\frac{5t^6\left(t+\frac1t\right)}{t^2-1} =\frac{5t^5(t^2+1)}{t^2-1}.$$ And $$y=t^5.$$ --- 6. **Substitute into the differential equation** Given $$(x^2-1)\frac{d^2y}{dx^2}+\lambda x\frac{dy}{dx}+ky=0.$$ Substitute the expressions: $$\frac{10t^5(2t^2-3)}{t^2-1}+\lambda\cdot \frac{5t^5(t^2+1)}{t^2-1}+kt^5=0.$$ Factor out $t^5$: $$t^5\left[\frac{10(2t^2-3)+5\lambda(t^2+1)}{t^2-1}+k\right]=0.$$ Multiply by $(t^2-1)$: $$10(2t^2-3)+5\lambda(t^2+1)+k(t^2-1)=0.$$ Expand: $$20t^2-30+5\lambda t^2+5\lambda+kt^2-k=0.$$ Group coefficients: $$(20+5\lambda+k)t^2+(-30+5\lambda-k)=0.$$ Since this holds for all $t$, both coefficients must be zero: $$20+5\lambda+k=0 \quad ...(1)$$ $$-30+5\lambda-k=0 \quad ...(2)$$ Add (1) and (2): $$10\lambda-10=0 \implies \lambda=1.$$ Then from (1): $$20+5(1)+k=0 \implies 25+k=0 \implies k=-25.$$ Thus, $$\lambda+k=1+(-25)=-24.$$ --- 7. **Check options** The value is $$-24.$$ So the correct option is: **B: $-24$** --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer is also **B**. Hence they agree.

Differential Equations question

Differential Equations question

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