1. **Problem a:** Solve the differential equation $$y'=1+e^{y-x+5}$$. Step 1: Rewrite the equation as $$\frac{dy}{dx}=1+e^{y-x+5}$$. Step 2: Let us try to separate variables. Rearrange terms: $$\frac{dy}{1+e^{y-x+5}}=dx$$. Step 3: Substitute $$u = y - x + 5$$ so that $$du = dy - dx \Rightarrow dy = du + dx$$. Step 4: Substitute into the separated form: $$\frac{du+dx}{1+e^{u}} = dx \Rightarrow \frac{du}{1+e^{u}} + \frac{dx}{1+e^{u}} = dx$$. Step 5: Rearranged to isolate $$du$$ term: $$\frac{du}{1+e^{u}} = dx - \frac{dx}{1+e^{u}} = dx\left(1 - \frac{1}{1+e^{u}}\right) = dx\frac{e^{u}}{1+e^{u}}$$. Step 6: So, $$\frac{du}{1+e^{u}} = \frac{e^{u}}{1+e^{u}} dx \Rightarrow du = e^{u} dx$$. Step 7: But since $$u=y-x+5$$, this rearrangement leads to an inconsistency in direct separation, so approach by rewriting original equation as: $$y' - 1 = e^{y-x+5}$$. Step 8: Let $$v = y - x + 5$$, then $$y' = \frac{dy}{dx} = v' + 1$$ (since $$v = y - x + 5$$, so $$v' = y' - 1$$). Step 9: Substitute into equation: $$y' = 1 + e^{v} \Rightarrow v' + 1 = 1 + e^{v} \Rightarrow v' = e^{v}$$. Step 10: Separate variables: $$\frac{dv}{e^{v}} = dx$$. Step 11: Simplify left side: $$e^{-v} dv = dx$$. Step 12: Integrate both sides: $$\int e^{-v} dv = \int dx$$. Step 13: The integral of $$e^{-v}$$ is $$-e^{-v}$$, so $$-e^{-v} = x + C$$. Step 14: Rearrange for $$v$$: $$e^{-v} = -x - C$$. Step 15: Since $$v=y - x + 5$$, $$e^{-y+x-5} = -x - C$$. Step 16: Taking logarithm and solving explicitly for $$y$$ is possible but implicit form suffices: $$e^{x - y - 5} = -x - C$$. --- 2. **Problem b:** Solve $$y'' = 1 + (y')^2$$ with initial conditions $$y(0)=0$$ and $$y'(0)=0$$. Step 1: Let $$p = y'$$ so that $$p' = y''$$. Step 2: Rewrite the ODE as $$p' = 1 + p^2$$. Step 3: Separate variables: $$\frac{dp}{1 + p^2} = dx$$. Step 4: Integrate both sides: $$\int \frac{dp}{1+p^2} = \int dx \Rightarrow \arctan p = x + C$$. Step 5: Use initial condition $$p(0) = 0$$: $$\arctan 0 = 0 = 0 + C \Rightarrow C=0$$. Step 6: Therefore, $$\arctan p = x \Rightarrow p = \tan x$$. Step 7: Recall $$p = y'$$, so $$y' = \tan x$$. Step 8: Integrate: $$y = \int \tan x \, dx = -\ln |\cos x| + C_1$$. Step 9: Apply $$y(0) = 0$$: $$0 = -\ln 1 + C_1 \Rightarrow C_1 = 0$$. Step 10: Final solution: $$y = -\ln |\cos x|$$. --- 3. **Problem c:** Consider the equation $$y^{(5)} x + y^{(4)} + \sin x = 0$$. Step 1: Identify terms with derivatives: - $$y^{(5)}$$ is the 5th derivative, - $$y^{(4)}$$ is the 4th derivative. Step 2: Note the presence of $$x$$ multiplying the 5th derivative. Step 3: An equation is linear if the dependent variable $$y$$ and all its derivatives appear to the first power and are not multiplied or composed with each other. Step 4: Here, the equation can be rewritten in form: $$x y^{(5)} + y^{(4)} = - \sin x$$, which is linear in $$y$$ and its derivatives since: - No powers other than 1, - No products like $$y y'$$ or nonlinear functions of $$y$$ or its derivatives, - Coefficients (like $$x$$) depend only on the independent variable $$x$$. Step 5: Hence, this is a linear differential equation with variable coefficients. **Final answers:** a) Implicit solution: $$-e^{- (y - x + 5)} = x + C$$. b) Explicit solution: $$y = -\ln |\cos x|$$. c) The equation $$y^{(5)} x + y^{(4)} + \sin x = 0$$ is a linear differential equation with variable coefficients.