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lovelymultani

  • one year ago

Use separation of variables to solve the initial value problem. dy/dx = e^x/y , y(0)=-4

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  1. dumbcow
    • one year ago
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    \[\rightarrow \int\limits_{}^{} y dy = \int\limits_{}^{} e^{x} dx\] \[\frac{y^{2}}{2} = e^{x} + C\] \[y = \pm \sqrt{2e^{x} +C}\] plug in inital point to find C \[-4 = -\sqrt{2+C}\] C = 14

  2. lovelymultani
    • one year ago
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    so the final answer would be \[y=\pm \sqrt{2e^{x}+14}\] ?

  3. surjithayer
    • one year ago
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    \[\frac{ dy }{dx }=\frac{ e ^{x} }{ y }\] \[y dy=e ^{x} dx\] \[integrating,\] \[\int\limits y dy=\int\limits e ^{x} dx+c\] \[\frac{ y ^{2} }{2}=e ^{x}+c\] when x=0, y=-4 \[\frac{ \left( -4 \right)^{2} }{ 2 }=e ^{0}+c, 8=1+c, c=7\] \[\frac{ y ^{2} }{ 2}=e ^{x}+7\] \[y ^{2}=2 e ^{x}+14\]

  4. dumbcow
    • one year ago
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    final answer \[y = -\sqrt{2e^{x}+14} or -\sqrt{2} \sqrt{e^{x} +7}\]

  5. sweetey
    • one year ago
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    pls help solove this my assignment.i attach my assignment

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  6. surjithayer
    • one year ago
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    \[\lim_{x \rightarrow 2}\frac{ x ^{2}-4 }{ x-2 }=\lim_{x \rightarrow 2}\frac{ \left( x+2 \right)\left( x-2 \right) }{ x-2 }\] \[\lim_{x \rightarrow 2 }\left( x+2 \right)=2+2=4\] for( 2) y=1/x \[\lim_{\delta x \rightarrow 0}\frac{ \delta y }{\delta x }=\lim_{\delta x \rightarrow 0}\frac{ y+ \delta y-y }{\delta x }\] \[=\lim_{\delta x \rightarrow 0}\frac{ \frac{ 1 }{ x+ \delta x }-\frac{ 1 }{x } }{ \delta x }\] \[=\lim_{\delta x \rightarrow0}\frac{ \frac{ x-x-\delta x }{ x \left( x+\delta x \right) } }{ \delta x }\] \[=\lim_{\delta x \rightarrow 0}\frac{ -\delta x }{\delta x \left( x \right)\left( x+\delta x \right) }\] \[=\lim_{\delta x \rightarrow 0}\frac{ -1 }{ x \left( x+\delta x \right) }=\frac{ -1 }{ x \left( x+0 \right) }\] \[=\frac{ -1 }{ x ^{2} }\] for (2)1 \[y=\left(x ^{4}+3x ^{2} \right)\left( 5x ^{2} +10\right)\] \[\frac{ dy }{dx }=\left( x ^{4}+3x ^{2} \right)\left( 10x \right)+\left( 4x ^{3}+6x \right)\left(5x ^{2 } +10\right)\] \[=10 x ^{5}+30x ^{3}+20x ^{5}+40x ^{3}+30x ^{3}+60x\] =\[30x ^{5}+100x ^{3}+60x\] \[y=\left(x ^{3}+\left( x ^{2} \right)^{\frac{ 1 }{3 }} \right)\left(x ^{4} +3x ^{2}\right)\left( 5x ^{2}+10 \right)\] \[\frac{ dy }{ dx }=\left( 3x ^{2}+\frac{ 2 }{ 3 }x ^{\frac{ -1 }{ 3 }} \right)\left( x ^{4}+3x ^{2 } \right)\left( 5x ^{2 }+10 \right)\]+\[\left( x ^{3}+\left( x ^{2} \right)\frac{ 1 }{3 }\right)\frac{ d }{ dx }\left\{ \left( x ^{4}+3x ^{2} \right)\left( 5x ^{2 }+10 \right) \right\}\] now you can solve. (solved above)

  7. surjithayer
    • one year ago
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    for (3) \[\left[\begin{matrix}-3 & x \\2y& 0\end{matrix}\right]+\left[\begin{matrix}4 &6 \\-3 & 1\end{matrix}\right]=\left[\begin{matrix}1 & 7 \\ -5 & 1\end{matrix}\right]\] \[\left[\begin{matrix}-3+4 & x+6 \\ 2y-3&0+1\end{matrix}\right]=\left[\begin{matrix}1 &7 \\-5 & 1\end{matrix}\right]\] \[\left[\begin{matrix}1 & x+6 \\2y-3 & 1\end{matrix}\right]=\left[\begin{matrix}1 & 7 \\ -5 & 1\end{matrix}\right]\] x+6=7,x=7-6=1 2y-3=-5, 2y=-5+3=-2 y=-1

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