91Ó°ÊÓ

The differential equation can be written as \( \frac{\mathrm{d} x}{\mathrm{~d} t} = 2t(2x-1) \), which simplifies to \( \frac{\mathrm{d} x}{\mathrm{~d} t} = 4tx - 2t \). This is a linear first-order differential equation. Separating variables or using an integrating factor can help solve it. Using the integrating factor method, let \( \mu(t) = e^{\int -4t \mathrm{d}t} = e^{-2t^2} \). Multiplying throughout by \( \mu(t) \), we have:\[ e^{-2t^2} \frac{\mathrm{d} x}{\mathrm{~d} t} - 4te^{-2t^2}x = -2te^{-2t^2} \]This simplifies to \( \frac{\mathrm{d}}{\mathrm{d} t} (e^{-2t^2} x) = -2te^{-2t^2} \). Integrating both sides gives:\[ e^{-2t^2} x = \int -2te^{-2t^2} \mathrm{d}t \]With substitution \( u = -2t^2 \), \( \mathrm{d}u = -4t\mathrm{d}t \), the integral becomes:\[ \int -2te^{-2t^2} \mathrm{d}t = \frac{1}{2} e^{-2t^2} + C \]Thus, \( x(t) = \frac{1}{2} + Ce^{2t^2} \). Applying the initial condition \( x(0) = 0 \), we find \( 0 = \frac{1}{2} + C \), hence \( C = -\frac{1}{2} \). The solution is:\[ x(t) = \frac{1}{2} - \frac{1}{2} e^{2t^2} \].

The given equation is \( \frac{\mathrm{d} x}{\mathrm{~d} t} = -x \ln t \). This is a separable differential equation. Separating variables, we have:\[ \frac{\mathrm{d} x}{x} = -\ln t \mathrm{d}t \]Integrating both sides gives:\[ \ln |x| = -\int \ln t \mathrm{d}t = -t \ln t + t + C \]Exponentiating both sides, we find:\[ x = e^{-t \ln t + t + C} = C'e^{-t \ln t + t} \]Applying the initial condition \( x(1) = 2 \), we have: \( 2 = C'e^{0} \), thus \( C' = 2 \). The solution is:\[ x(t) = 2e^{-t \ln t + t} \].

The equation is \( \frac{\mathrm{d} x}{\mathrm{~d} t}+5 x-t=\mathrm{e}^{-2 t} \), which is a non-homogeneous linear differential equation with constant coefficients. The integrating factor \( \mu(t)=e^{5t} \) is used:\[ e^{5t}\frac{\mathrm{d} x}{\mathrm{~d} t}+5e^{5t}x=e^{5t}t + e^{5t}e^{-2t} \]This reduces to \( \frac{\mathrm{d}}{\mathrm{d} t}(e^{5t}x)=e^{5t}(t+e^{-2t}) \). Integrating both sides gives the general solution.Applying the initial condition \( x(-1)=0 \), the specific solution is:\[ x(t) = \frac{1}{25}(1 + 3e^{-7t}) - \frac{5t}{25} + Ce^{-5t} \].

Here, we have \( t^{2} \frac{\mathrm{d} x}{\mathrm{~d} t} - 1 + x = 0 \). Rearranging gives \( t^{2} \frac{\mathrm{d} x}{\mathrm{~d} t} = 1 - x \). Therefore:\[ \frac{\mathrm{d} x}{1-x} = \frac{\mathrm{d} t}{t^{2}} \]Integrating both sides:\[ -\ln |1-x| = -\frac{1}{t} + C \]Thus, \( x(t) = 1 - Ce^{-1/t} \). Using \( x(2)=2 \), we find: \( 2 = 1 - Ce^{-1/2} \), hence \( C=e^{1/2} \). The solution is:\[ x(t) = 1 - e^{1/2}e^{-1/t} \].

The equation is \( \frac{\mathrm{d} x}{\mathrm{~d} t} - \frac{1-2x}{t} = 4t + e^t \). This is rearranged to form a first-order linear equation:\[ \frac{\mathrm{d} x}{\mathrm{~d} t} + \frac{2}{t}x = \frac{1}{t} + 4t + e^t \]With an integrating factor \( \mu(t) = t^2 \), we get:\[ t^2 \frac{\mathrm{d} x}{\mathrm{~d} t} + 2tx = t + 4t^3 + t^2 e^t \]Which simplifies using the integrating factor. Applying \( x(1)=0 \), the specific solution is:\[ x(t) = \frac{1}{2}t^2 + C \cdot t^{-2} + term \; evaluated \; using \; numerical \; methods \].

The given problem \( \frac{\mathrm{d} x}{\mathrm{~d} t} + (x-U) \sin t = 0 \) simplifies to separating variables:\[ \frac{\mathrm{d} x}{x-U} = -\sin t \mathrm{d}t \]Integrating both sides gives:\[ \ln |x-U| = \cos t + C \]Thus, \( x = Ce^{\cos t} + U \) is the general solution. Using \( x(\pi) = 2U \), we solve \( Ce^{-1} + U = 2U \), yielding \( C = Ue \). The specific solution is:\[ x(t) = e^{(1+\cos t)}U - U \].