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Solve the equation e2x dy/dx 1 given that y = 5 when x = 0 ​

1 Answer

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Answer:

The equation is


{ \underline{ \sf{y = - 2 {e}^( - 2x) + 7 }}}

Explanation:


{ \sf{ {e}^(2x) (dy)/(dx) = 1}} \\ \\ { \sf{dy = {e}^( - 2x) dx}}

integrate:


{ \sf{ \int dy = \int { - e}^(2x) dx}} \\ { \sf{y = - 2 {e}^( - 2x) + c}}

c is a constant.

when y is 5, x is 0:


{ \sf{y = { - 2e}^( - 2x) + c}} \\ { \sf{5 = - 2 {e}^(( - 2 * 0)) + c }} \\ { \sf{5 = - 2 {e}^(0) + c }} \\ { \sf{5 = ( - 2 * 1) + c}} \\ { \sf{5 = - 2 + c}} \\ { \sf{c = 7}}

therefore, equation:


y = - 2 {e}^( - 2x) + 7

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