Find y(e) for \(\rm x^2 {dy\over dx}+4xy=4{\ln⁡x\over x^3}\), and y

1.	Find y(e) for \(\rm x^2 {dy\over dx}+4xy=4{\ln⁡x\over x^3}\), and y(1) = 11. \(\rm 2\over e^3\)2. \(\rm 3\over e^4\)3. \(\rm 4\over e^2\)4. \(\rm 5\over e^5\)
Answer» Correct Answer - Option 2 : \(\rm 3\over e^4\) Concept: In first order linear differential equation; \(\rm {dy\over dx}+Py=Q\), where P and Q are function of x Integrating factor (IF) = e^{∫ P dx} y × (IF) = ∫ Q(IF) dx Calculation: Linear differential equation is of first order \(\rm x^2 {dy\over dx}+4xy=4{\ln⁡x\over x^3}\) ⇒ \(\rm {dy\over dx}+4{y\over x}=4{\ln⁡x\over x^5}\) IF = e^{∫ \(\rm 4\over x\) dx} ⇒ IF = e^{4 ln x} ⇒ IF = x⁴ Now, y × (IF) = ∫ Q (IF) dx ⇒ y × x⁴ = ∫ 4 \(\rm \ln x\over x^5\) × x⁴ dx ⇒ yx⁴ = ∫ 4 \(\rm \ln x\over x\) dx Integrating, ⇒ yx⁴ = 2 (ln x)² + c (where c is integration constant) Given y(1) = 1 ⇒ (1)(1)⁴ = 2 (ln 1)² + c ⇒ c = 1 ∴ yx⁴ = 2 (ln x)2 + 1 For y(e) y(e)⁴ = 2 (ln e)² +1 ⇒ y(e⁴) = 3 ⇒ y = \(\boldsymbol{\rm 3\over e^4}\)

Find y(e) for \(\rm x^2 {dy\over dx}+4xy=4{\ln⁡x\over x^3}\), and y(1) = 11. \(\rm 2\over e^3\)2. \(\rm 3\over e^4\)3. \(\rm 4\over e^2\)4. \(\rm 5\over e^5\)

Answer» Correct Answer - Option 2 : \(\rm 3\over e^4\)

Concept:

In first order linear differential equation;

\(\rm {dy\over dx}+Py=Q\), where P and Q are function of x

Integrating factor (IF) = e^{∫ P dx}

y × (IF) = ∫ Q(IF) dx

Calculation:

Linear differential equation is of first order

\(\rm x^2 {dy\over dx}+4xy=4{\ln⁡x\over x^3}\)

⇒ \(\rm {dy\over dx}+4{y\over x}=4{\ln⁡x\over x^5}\)

IF = e^{∫ \(\rm 4\over x\) dx}

⇒ IF = e^{4 ln x}

⇒ IF = x⁴

Now, y × (IF) = ∫ Q (IF) dx

⇒ y × x⁴ = ∫ 4 \(\rm \ln x\over x^5\) × x⁴ dx

⇒ yx⁴ = ∫ 4 \(\rm \ln x\over x\) dx

Integrating,

⇒ yx⁴ = 2 (ln x)² + c (where c is integration constant)

Given y(1) = 1

⇒ (1)(1)⁴ = 2 (ln 1)² + c

⇒ c = 1

∴ yx⁴ = 2 (ln x)2 + 1

For y(e)

y(e)⁴ = 2 (ln e)² +1

⇒ y(e⁴) = 3

⇒ y = \(\boldsymbol{\rm 3\over e^4}\)