# Tutor profile: Joe R.

## Questions

### Subject: Linear Algebra

Find the eigenvalues and eigenvectors of the following matrix. $(A = \begin{pmatrix} 2 & 7 \\ -1 & -6 \end{pmatrix}$)

Recall that he eigenvalues are the numbers $$\lambda$$ so that $$A - \lambda I$$ is singular. In other words, they are the solutions to $$\det(A - \lambda I) = 0$$. Since $(A - \lambda I = \begin{pmatrix} 2 & 7 \\ -1 & -6 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 - \lambda & 7 \\ -1 & -6-\lambda \end{pmatrix}$) we want to solve $$(2-\lambda)(-6-\lambda) - (7)(-1) = \lambda^2 +4 \lambda -5 = 0$$. This quadratic in $$\lambda$$ factors as $$(\lambda+5)(\lambda - 1)$$ so the two eigenvalues are $$\lambda_1 = -5$$ and $$\lambda_2 = 1$$. Now let us find the corresponding eigenvectors. For $$\lambda_1 = -5$$, we want to solve $$(A+5I)\bf{v} = \bf{0}$$. In other words, we seek a solution to $( \begin{pmatrix} 7 & 7 \\ -1 & -1 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}.$) The augmented matrix for this system of equations is $(\begin{pmatrix} 7 & 7 & 0 \\-1 & -1 & 0 \end{pmatrix}$) and recall that we can row-reduce this augmented matrix to find the solution. Since $(\begin{pmatrix} 7 & 7 & 0 \\-1 & -1 & 0 \end{pmatrix}$) is row-equivalent to $(\begin{pmatrix} 1 & 1 & 0 \\0 & 0 & 0 \end{pmatrix},$) the solution is given by any multiple of the vector $$v = \begin{pmatrix} 1 \\-1 \end{pmatrix}$$ and so is an eigenvector corresponding to $$\lambda_1 = -5$$. Similarly for $$\lambda_2 = 1$$, we find the augmented matrix to be $(\begin{pmatrix} 1 & 7 & 0 \\ -1 & -7 & 0 \end{pmatrix}$) which is equivalent to $(\begin{pmatrix} 1 & 7 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$) Therefore we can choose our eigenvector to be $$\bf{v} = \begin{pmatrix} -7 \\ 1 \end{pmatrix}$$.

### Subject: Calculus

Evaluate the following integral. $(\int e^x \sqrt{e^{2x} + 1} \ dx$)

First make the substitution $$t = e^x$$ and $$dt = e^x dx$$. Then the integral becomes $(\int \sqrt{t^2 + 1} \ dt.$) This integral can be computed using the trig substitution $$t = \tan\theta$$. Applying this transformation, the integral becomes $(\int \sqrt{\tan^2 \theta + 1} \sec^2 \theta \ d\theta.$) Using the identity $$\tan ^2 \theta + 1 = \sec^2 \theta$$, we can rewrite the integral as $(\int \sec^3 \theta \ d\theta.$) Now apply integration by parts with $$u = \sec \theta$$ and $$dv = \sec^2 \theta \ d\theta$$. Then we get $(\int \sec^3 \theta \ d\theta = \sec \theta \tan \theta - \int \tan^2 \theta \sec \theta \ d\theta.$) Again using the identity $$\tan ^2 \theta + 1 = \sec^2 \theta$$, we can write $(\int \sec^3 \theta \ d\theta = \sec \theta \tan \theta - \int \sec^3 \theta - \sec\theta \ d\theta$) so that $(\int 2 \sec^3 \theta \ d\theta = \sec \theta \tan \theta +\int \sec\theta \ d\theta$) and so $(\int \sec^3 \theta \ d\theta = \frac{1}{2}\left(\sec \theta \tan \theta + \ln(|\sec\theta + \tan\theta|) \right) + C$). Now since $$t = \tan \theta$$ we have $$\sqrt{t^2 + 1} = \sec \theta$$, so transforming the variables back from $$\theta$$ to $t$, the answer becomes $(\int \sqrt{t^2 + 1} \ dt = \frac{1}{2}(t \sqrt{t^2 + 1} + \ln(|t + \sqrt{t^2 +1 }|) + C$) and finally converting back to $$x$$ from $$t$$, we get $(\int e^x \sqrt{e^{2x} + 1} \ dx = \frac{1}{2}(e^x \sqrt{e^{2x} + 1} + \ln(|e^x + \sqrt{e^{2x} +1 }|) + C.$)

### Subject: Differential Equations

Solve the following IVP. $$y^{\prime \prime} -4y^{\prime} +9y = 0 \ \ \ y(0) = 0 \ \ \ y^{\prime}(0) = -8$$

Since the differential equation is second-order, homogenous, and has constant coefficients, we first look at the characteristic equation. In this case, the characteristic equation is $(r^2 - 4r +9 = 0$$) which has roots $(r = \frac{4 \pm \sqrt{-20}}{2} = 2 \pm \sqrt{5} i$). Since the roots are complex with non-zero imaginary part, we know the general solution is $(y(t) = c_1 e^{2t}\cos(\sqrt{5}t) + c_2 e^{2t}\sin(\sqrt{5}t).$) Applying the first initial condition gives the equation $(0 = y(0) = c_1 $) so the solution reduces to $(y(t) = c_2e^{2t} \sin(\sqrt{5}t).$) The second initial condition yields $( -8 = y^{\prime}(0) = \sqrt{5}c_2$) so that $(c_2 = -\frac{8}{\sqrt{5}}$) and therefore the solution to the IVP is $(y(t) = -\frac{8}{\sqrt{5}} e^{2t} \sin(\sqrt{5}t).$)

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