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Digvijay P.
Assistant professor for 3 years, Senior research fellow at IIT Roorkee
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Linear Algebra
TutorMe
Question:

Provide the geometric interpretation for eigenvectors of a real matrix.

Digvijay P.
Answer:

The eigenvectors of a real matrix form a basis. It can also be said that the eigenvectors are the vectors which when transformed by the matrix, results in stretching of the eigenvector. The amount of stretching is given by the eigenvalue.

Civil Engineering
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Question:

Enlist the different branches of civil engineering.

Digvijay P.
Answer:

Structural engineering Geotechnical engineering Water resources engineering surveying Environmental engineering Construction technology Transportation engineering Hydrology

Differential Equations
TutorMe
Question:

The governing differential equation for the harmonic oscillator is given by : $(\dfrac{d^2u}{dt^2} + \omega^2u = 0$) Where $$u$$ is displacement of the oscillator with time $$t$$. Obtain the solution for $$u(t)$$ if $(u(t=0) = u_0 \hspace{10pt} and \hspace{10pt}\dot u(t=0) = \dot u_0$).

Digvijay P.
Answer:

The solution of such equation will be of the form $( u(t) = Ae^{\lambda t}$) $( \therefore \dot u(t) = A\lambda e^{\lambda t}$) $( \ddot u(t) = A\lambda^2 e^{\lambda t}$) Where number of dots above variable $$u$$ indicated the order of derivative of $$u$$ w.r.t. $$t$$. $$A$$ and $$\lambda$$ are constants. Substituting the above expressions in the differential equation gives : $(A\lambda^2e^{\lambda t} + \omega^2 Ae^{\lambda t} = 0$) As $$A$$ is an arbitrary constant and $$e^{\lambda t} \neq 0 \hspace{10pt}\forall \hspace{10pt}t$$. Dividing the whole expression by $$Ae^{\lambda t}$$, we get $(\lambda^2 + \omega^2 = 0$) $(\lambda = \pm i\omega$) Hence the general solution can be written as $( u(t) = A_1e^{+i\omega t} + A_2 e^{-i\omega t}$) The values for constants can be obtained from initial conditions, @$$t=0 \hspace{10pt} u = u_0 \implies \hspace{10pt} u_0 = A_1 + A_2$$ @$$t=0 \hspace{10pt} \dot u = \dot u_0 \implies \hspace{10pt} \dot u_0 = i\omega(A_1-A_2)$$ Solving for $$A_1$$ and $$A_2$$ gives $(A_1 = \dfrac{1}{2}[u_0 + \dfrac{\dot u_0}{i\omega}]$) $(A_2 = \dfrac{1}{2}[u_0 - \dfrac{\dot u_0}{i\omega}]$) Substitute in $$u(t)$$ and using Euler's identity $( e^{\pm i\theta} = cos(\theta) \pm i sin(\theta)$) $( u(t) = u_0 cos(\omega t) + \dfrac{\dot u_0}{\omega} sin(\omega t)$)

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