QUESTIONS 1) Find the general solution of the following differential equation (30 points) y''+25y=cos(5x) 2) Solve the initial value problem by using Laplace transform (30 points) (d^2y)/(dt^2)+6(dy)/(dt)-7y=0 y(0)=1, y'(0)=-2 3) Find the power series solutions in powers of x in the following differential equation (40 points) (d^2y)/(dx^2)+3x(dy)/(dx)-4y=0

1) To find the general solution of the differential equation $y''+25y=\cos(5x)$, we first find the complementary solution by solving the homogeneous equation $y''+25y=0$. The characteristic equation is $r^2+25=0$, which has complex roots $r=\pm 5i$. Therefore, the complementary solution is $y_c(x)=c_1\cos(5x)+c_2\sin(5x)$.<br /><br />Next, we find a particular solution of the form $y_p(x)=A\cos(5x)+B\sin(5x)$. Substituting this into the original equation, we get $-25A\cos(5x)-25B\sin(5x)+25A\cos(5x)+25B\sin(5x)=\cos(5x)$. Simplifying, we get $-25B\sin(5x)=\cos(5x)$, which implies $B=-\frac{1}{25}$. Therefore, the particular solution is $y_p(x)=-\frac{1}{25}\sin(5x)$.<br /><br />The general solution is $y(x)=y_c(x)+y_p(x)=c_1\cos(5x)+c_2\sin(5x)-\frac{1}{25}\sin(5x)$.<br /><br />2) To solve the initial value problem using Laplace transform, we take the Laplace transform of both sides of the differential equation $\frac{d^2y}{dt^2}+6\frac{dy}{dt}-7y=0$. Applying the Laplace transform to each term, we get $s^2Y(s)-sy(0)-y'(0)+6(sY(s)-y(0))-7Y(s)=0$, where $Y(s)$ is the Laplace transform of $y(t)$.<br /><br />Substituting the initial conditions $y(0)=1$ and $y'(0)=-2$, we get $s^2Y(s)-s-2+6(sY(s)-1)-7Y(s)=0$. Simplifying, we get $(s^2+6s-7)Y(s)=s+8$. Solving for $Y(s)$, we get $Y(s)=\frac{s+8}{s^2+6s-7}$.<br /><br />To find $y(t)$, we take the inverse Laplace transform of $Y(s)$. Using partial fraction decomposition, we get $Y(s)=\frac{1}{s-1}+\frac{7}{s+7}$. Therefore, $y(t)=e^t+7e^{-7t}$.<br /><br />3) To find the power series solutions of the differential equation $\frac{d^2y}{dx^2}+3x\frac{dy}{dx}-4y=0$, we assume a power series solution of the form $y(x)=\sum_{n=0}^{\infty}a_nx^n$. Substituting this into the differential equation, we get $\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}+3\sum_{n=0}^{\infty}na_nx^{n+1}-4\sum_{n=0}^{\infty}a_nx^n=0$.<br /><br />To simplify the equation, we shift the indices of the first two sums by 2 and 1, respectively. This gives us $\sum_{n=2}^{\infty}(n-2)(n-1)a_{n-2}x^{n-2}+3\sum_{n=1}^{\infty}na_{n-1}x^{n}