Question

\(\frac{dy}{dx}\) + \(\frac{5}{x(1+x^5)}\)y = \(\frac{(1+x^5)^2}{x^7}\) If y(1) = 2, then the value of y(2) is:

• A. \(\frac{693}{128}\)
• B. \(\frac{697}{128}\)
• C. \(\frac{637}{128}\)
• D. \(\frac{627}{128}\)

Answer 1

The Correct Option is A

Step 1: Determine the Integrating Factor (I.F.)

The integrating factor is given by:

\[ \text{I.F.} = e^{\int \frac{1}{x^5 + 1} \, dx}. \]

Let \( t = 1 + x^5 \), so that \( dt = 5x^4 \, dx \). Substituting, the integrating factor simplifies to:

\[ \text{I.F.} = e^{-\ln t} = \frac{1}{t} = \frac{1}{1 + x^5}. \]

Step 2: Solve for \( y \)

Multiply through by the integrating factor:

\[ y \cdot \frac{x^5}{1 + x^5} = \int \frac{x^4}{1 + x^5} \, dx. \]

Using substitution \( t = 1 + x^5 \), this becomes:

\[ \int \frac{x^4}{1 + x^5} \, dx = \int \frac{1}{t} \, dt = \ln t = \ln(1 + x^5). \]

Thus, the solution is:

\[ y \cdot \frac{x^5}{1 + x^5} = \ln(1 + x^5) + C. \]

Step 3: Solve for \( y \)

Rewriting the equation for \( y \):

\[ y = \frac{\ln(1 + x^5) + C}{\frac{x^5}{1 + x^5}} = \frac{\ln(1 + x^5) + C}{x^5}. \]

Step 4: Apply Initial Conditions

Given \( x = 1 \) and \( y = 2 \):

\[ 2 \cdot \frac{1}{2} = \ln(1 + 1) + C. \]

Substituting \( \ln(2) \) for \( \ln(1 + 1) \):

\[ 1 = \ln 2 + C \implies C = 1 - \ln 2. \]

Step 5: Compute \( y \) for \( x = 2 \)

Substitute \( x = 2 \) into the solution for \( y \):

\[ y = \frac{\ln(1 + 2^5) + C}{2^5} = \frac{\ln(33) + 1 - \ln 2}{32}. \]

Combining terms:

\[ y = \frac{\ln 33 - \ln 2 + 1}{32}. \]

Final Answer:

\( y = \frac{693}{128} \).

Answer 2

The correct option is (A): \(\frac{693}{128}\)
\(I.F=e^{\int\frac{5}{x(1+x^5)}dx}=e^{\int\frac{5x^{-6}}{(x^{-5}+1)}dx}\)
\(=e^{-ln{}(x^{-5}+1)}=\frac{1}{x^{-5}+1}=\frac{x^5}{x^5+1}\)
\(y.\frac{x^5}{x^5+1}=\int\frac{(1+x^5)^2}{x^7}.\frac{x^5}{(1+x^5)}dx\)
\(=\int\frac{(1+x^5)}{x^2}dx\)
\(=\frac{-1}{x}+\frac{x^4}{4}+C\)
\(y(1)=2⇒2(\frac{1}{2})=-1+\frac{1}{4}+C\)
\(⇒C=\frac{7}{4}\)
Put x=2
\(⇒y(\frac{32}{33})=\frac{-1}{2}+4+\frac{7}{4}\)
\(⇒ y=\frac{693}{128}\)