Solve the initial value problem: r′(t) = 5i+5etj+(3et+3tet)k and r(0)=3i+5j−2k

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Solution:The given vector function:r^{\prime}(t)=5 i+5 e^{t} j+\left(3 e^{t}+3 t e^{t}\right) jand initial condition \Omega(0)=3 i+5 j-2 kintegration\begin{aligned}r(t) & =\int r^{\prime}(t) d t \\& =\int\left(5 i+5 e^{t} j+\left(3 e^{t}+3 t e^{t}\right) j k\right) d t \\& =5 t i+5 e^{t} j+\left(3 e^{t}+3\left(\int t e^{t}\right) j\right)+c \\& =5 t i+5 e^{t} j+\left(3 e^{t}+3\left[t e^{t}-e^{t}\right]\right) j+c \\& =5 t i+5 e^{t} j+\left(3 t e^{t}\right) j+c\end{aligned}r(t)=5 t i+5 e^{t} j+3 t e^{t} j+cNow when t=0 \quad \Omega(0)=3 i+5 j-2 k\begin{array}{l}3 i+5 j-2 k=5(0) i+5 e^{0} j+3(0) e^{0} j+c \\3 i+5 j-2 k=5 j+c \\c=3 i-2 k\end{array}Hence\begin{array}{r}r(t)=5 t i+5 e^{t} j+3 t e^{t} j+3 i-2 k \\\Omega(t)=(5 t+3) i+5 e^{t} j+\left(3 t e^{t}-2\right) \dot{k}\end{array} ...