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qila

  • 8 months ago

verify that y=(4e^3x) -2 is an explicit solution of differential equation y'-3y=6

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  1. esamalaa
    • 8 months ago
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    use the variable separation to solve this D E

  2. qila
    • 8 months ago
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    can you try to solve it and explain more about it?

  3. esamalaa
    • 8 months ago
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    i think that's the right solution

  4. VeritasVosLiberabit
    • 8 months ago
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    \[\frac{ dy }{ dx }=6+3y\] \[\frac{ dy }{ 6+3y }=dx\] \[\int\limits_{}^{}\frac{ dy }{ 6+3y }=\int\limits_{}^{}dx\] \[\frac{ 1 }{ 3 }\ln(6+3y)+C _{1}=x+C _{2}\]

  5. VeritasVosLiberabit
    • 8 months ago
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    \[\ln(6+3y)=3x+C\] \[e ^{\ln(6+3y)}=e ^{3x+C}\] \[6+3y=e ^{3x}e ^{C}\] \[3y=Ce ^{3x}-6\] \[y=Ce ^{3x}-2\]

  6. VeritasVosLiberabit
    • 8 months ago
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    I think you need to know an intial value y_0 to find C = 4.

  7. UnkleRhaukus
    • 8 months ago
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    This question does not ask you to solve the DE, it asks to verify that a given solution solves the DE.

  8. UnkleRhaukus
    • 8 months ago
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    The function \[y(x)=4e^{3x} -2\] its derivative\[y'(x) = 12e^{3x}\] The differential equation \[\qquad y'-3y\qquad \quad =6\] Plugging the function and its derivative into the DE\[\begin{align} [12e^{3x}]-3[4e^{3x} -2]&=6\\ 12e^{3x}-12e^{3x} +6&=6\\ 6&=6\\ 0&=0 \end{align}\](a true statement) Hence \(y=4e^{3x} -2\) is an explicit solution to the differential equation.

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