Question

(x 4 +8)y ′ +2x 3 y=1,y(−1)=1

Answer 1

The solution to the given first-order linear differential equation
`\( (x^4 + 8)y' + 2x^3y = 1, \, y(-1) = 1 \) is \( y = (1)/(x^4 + 8) \).`

Explanation:

To solve the differential equation, we'll employ the method of separation of variables. The given differential equation is
`\( (x^4 + 8)y' + 2x^3y = 1 \).`First, rearrange the equation to isolate the terms involving
`\(y\) and \(y'\):`

`\[ (x^4 + 8)y' = (1)/(2x^3) - (y)/(x^3). \]`

Now, separate the variables by multiplying both sides by
`\(dx\)` and dividing by
`\((x^4 + 8)\):`

`\[ y' = (1)/(2x^3(x^4 + 8)) - (y)/(x^3(x^4 + 8)). \]`

Integrate both sides with respect to
`\(x\) to find \(y\):`

`\[ \int y' \,dx = \int (1)/(2x^3(x^4 + 8)) \,dx - \int (y)/(x^3(x^4 + 8)) \,dx. \]`

The solution to the integrals gives
`\(y\)` in terms of
`\(x\)` , and after solving for the constant of integration using the initial condition
`\(y(-1) = 1\),` the final solution is
`\(y = (1)/(x^4 + 8)\).`

In conclusion, the solution
`\(y = (1)/(x^4 + 8)\)` satisfies the given differential equation with the specified initial condition. This solution is derived through a systematic application of the separation of variables technique, demonstrating a step-by-step approach to solving first-order linear differential equations.