`dy/dx = 5e^(-x/2)` Use integration to find a general solution to the differential equation

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 The given problem: ` (dy)/(dx) = 5e^(-x/2)`  is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: `N(y) dy= M(x)dx`

Cross-multiply `dx` to the other side, we get:

`dy= 5e^(-x/2)dx`

In this form, we may now proceed...

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 The given problem: ` (dy)/(dx) = 5e^(-x/2)`  is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: `N(y) dy= M(x)dx`

Cross-multiply `dx` to the other side, we get:

`dy= 5e^(-x/2)dx`

In this form, we may now proceed to direct integration on both sides:

`int dy= int 5e^(-x/2)dx`

For the left side, we apply basic integration property: `int (dy)=y` .

For the right side, we may apply u-substitution by letting: `u = -x/2` then `du =-1/2 dx` or `-2du= dx` .

Plug-in the values: `-x/2=u` and `dx=-2du` , we get:

`int 5e^(-x/2)dx=int 5e^(u)* (-2 du)`

                  ` =int -10e^(u)du`

Apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .

`int -10e^(u) du=(-10) int e^(u) du`

Apply basic integration formula for exponential function:

`(-10)int e^(u) du= -10e^(u)+C`

Plug-in` u=-x/2` on` -10e^(u)+C` , we get:

`int 5e^(-x/2) dx=-10e^(-x/2)+C`

Combining the results from both sides, we get the general solution of differential equation as:

`y=-10e^(-x/2)+C`

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