Question

Xy ′ +3y= x cos(x) ​ ,y(π)=0

Answer 1

The solution to the given initial value problem is
`\( y(x) = (1)/(4)x\sin(x) - (1)/(2)\cos(x) + (1)/(4)x\cos(x) \).`

Step-by-step explanation:

The given differential equation is a linear first-order ordinary differential equation (ODE) with an initial condition. To solve it, we can use the method of integrating factors. First, rearrange the equation to the standard form
`\( y' + P(x)y = Q(x) \)`, where
`\( P(x) \)` and
`\( Q(x) \)` are functions of
`\( x \)`. In this case,
`\( P(x) = 3 \) and \( Q(x) = x\cos(x) \)`.

The integrating factor
`\( \mu(x) \)` is found by exponentiating the integral of
`\( P(x) \)`, which gives
`\( \mu(x) = e^(\int 3 \,dx) = e^(3x) \)`. Multiply both sides of the differential equation by
`\( \mu(x) \)` to obtain the exact differential form
`\( e^(3x)y' + 3e^(3x)y = xe^(3x)\cos(x) \)`.

The left side is now the result of the product rule, and by applying it in reverse (integrating), we find
`\( e^(3x)y = (1)/(4)xe^(3x)\sin(x) - (1)/(2)e^(3x)\cos(x) + C \)`, where
`\( C \)` is the constant of integration.

Finally, solve for
`\( y \)` and apply the initial condition
`\( y(\pi) = 0 \)` to determine the value of the constant
`\( C \)`, yielding the final solution.