Solve the initial value problem given by \[ e^{-y} \frac{d y}{d x}+\left(x^{2}-\cos x\right)=-1 ; y(0)=-1 \]

To solve the initial value problem \[ e^{-y} \frac{dy}{dx} + \left(x^2 - \cos x\right) = -1 \] with the initial condition \(y(0) = -1\), we first rewrite the equation for clarity: \[ e^{-y} \frac{dy}{dx} = -1 - \left(x^2 - \cos x\right) \] This simplifies to: \[ e^{-y} \frac{dy}{dx} = -1 - x^2 + \cos x \] Next, we will separate the variables. We can rewrite it as: \[ \frac{dy}{dx} = \left(-1 - x^2 + \cos x\right) e^{y} \] Now, we separate the variables: \[ \frac{dy}{e^{y}} = \left(-1 - x^2 + \cos x\right) dx \] Integrating both sides: \[ \int \frac{dy}{e^y} = \int \left(-1 - x^2 + \cos x\right) dx \] The left integral can be solved as: \[ \int \frac{dy}{e^y} = -e^{-y} + C_1 \] For the right side, we integrate term by term: \[ \int \left(-1 - x^2 + \cos x\right) dx = -x - \frac{x^3}{3} + \sin x + C_2 \] Thus, we have: \[ -e^{-y} = -x - \frac{x^3}{3} + \sin x + C \] Where \(C = C_2 - C_1\). Rearranging gives: \[ e^{-y} = x + \frac{x^3}{3} - \sin x - C \] Now we need to apply the initial condition \(y(0) = -1\): Substituting \(x = 0\): \[ e^{-(-1)} = 0 + 0 - 0 - C \implies e = -C \implies C = -e \] Substituting \(C\) back into the equation gives: \[ e^{-y} = x + \frac{x^3}{3} - \sin x + e \] To isolate \(y\), we take the natural logarithm of both sides: \[ -y = \ln\left(x + \frac{x^3}{3} - \sin x + e\right) \] Thus: \[ y = -\ln\left(x + \frac{x^3}{3} - \sin x + e\right) \] This provides the general solution to the initial value problem. The final boxed solution looks like this: \[ y = -\ln\left(x + \frac{x^3}{3} - \sin x + e\right) \]