Question

solve :3x-y+logp=0
step to step
explain ​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The required solution is

$\displaystyle \sf{ 3{e}^{ - y} = {e}^{ - 3x} +c }$

Given :

The equation 3x - y + log p = 0

To find :

To solve the equation

Solution :

Step 1 of 2 :

Write down the given differential equation

Here the given differential equation is

3x - y + log p = 0

Step 2 of 2 :

Solve the equation

$\displaystyle \sf{ 3x - y + log p = 0 }$

$\displaystyle \sf{ \implies log p = y - 3x }$

$\displaystyle \sf{ \implies p = {e}^{y - 3x } }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = {e}^{y }. {e}^{ - 3x } }$

$\displaystyle \sf{ \implies {e}^{ - y}dy = {e}^{ - 3x} dx }$

On integration we get

$\displaystyle \sf{ \int {e}^{ - y}dy = \int {e}^{ - 3x} dx }$

$\displaystyle \sf{ \implies \frac{{e}^{ - y}}{ - 1} = \frac{{e}^{ - 3x} }{ - 3} + k }$

Where k is constant

$\displaystyle \sf{ \implies 3{e}^{ - y} = {e}^{ - 3x} - 3 k }$

$\displaystyle \sf{ \implies 3{e}^{ - y} = {e}^{ - 3x} +c }$

Where c = - 3k is a constant

Hence the required solution is

$\displaystyle \sf{ 3{e}^{ - y} = {e}^{ - 3x} +c }$

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