`xy + y' = 100x` Solve the differential equation

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The problem:` xy+y'=100x` is as first order differential equation that we can evaluate by applying variable separable differential equation:

`N(y)y'=M(x)`

`N(y)(dy)/(dx)=M(x)`

`N(y) dy=M(x) dx`

Apply direct integration:` intN(y) dy= int M(x) dx` to solve for the

 general solution of a differential equation.

Applying variable separable differential equation, we get:

...

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The problem:` xy+y'=100x` is as first order differential equation that we can evaluate by applying variable separable differential equation:

`N(y)y'=M(x)`

`N(y)(dy)/(dx)=M(x)`

`N(y) dy=M(x) dx`

Apply direct integration:` intN(y) dy= int M(x) dx` to solve for the

 general solution of a differential equation.

Applying variable separable differential equation, we get:

`xy+y'=100x`

`y' =100x-xy`

`y'=x(100-y)`

`(y')/(100-y)= x`

Let `y' =(dy)/(dx)` :

`((dy)/(dx))/(100-y)= x`

`(dy)/(100-y)= x dx`

Apply direct integration on both sides:

`int(dy)/(100-y)= int x dx`

For the left side, we consider u-substitution by letting:

`u= 100-y` then `du = -dy` or -`du=dy.`

The integral becomes:

`int(dy)/(100-y)=int(-du)/(u)`

Applying basic integration formula for logarithm:

`int(-du)/(u)= -ln|u|`

Plug-in `u = 100-y` on "`-ln|u|` " , we get:

`int(dy)/(100-y)=-ln|100-y|`

 

For the right side, we apply the Power Rule of integration: `int x^n dx = x^(n+1)/(n+1)+C`

 

 `int x* dx= x^(1+1)/(1+1)+C`

 

               ` = x^2/2+C`

 

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

`-ln|100-y|= x^2/2+C`

or 

`y =100- e^(-x^2/2-C)`

 `y = 100-Ce^(-x^2/2)

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