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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