How do you solve the differential equation `(dy)/dx=e^(y-x)sec(y)(1+x^2)`, where `y(0)=0` ?

By separating variables and integrating,

`int e^{-y}cosy dy=int (1+x^2)e^{-x} dx`

Integration by Parts 1
`u=e^{-y}" "dv=cosy dy`.
`du=-e^{-y} dy" "v=siny`.

Integration by Parts 2
`u=e^{-y}" "dv=siny dy`.
`du=-e^{-y}dy" "v=-cosy`

By Integration by Pats 1,

`(LHS)=e^{-y}siny+int e^{-y}siny dy`

by Integration by Parts 2,

`=e^{-y}siny-e^{-y}cosy-int e^{-y}cosy dy`

Since the last integral is the same as `(LHS)`, we have

`(LHS)=e^{-y}(siny-cosy)-(LHS)`

by adding `(LHS)`,

`Rightarrow2(LHS)=e^{-y}(siny-cosy)`

by dividing by `2`,

`Rightarrow (LHS)=e^{-y}/2(siny-cosy)`

Integration by Parts 3
`u=1+x^2" "dv=e^{-x}dx`
`du=2xdx" "v=-e^{-x}`

Integration by Parts 4
`u=2x" "dv=e^{-x}dx`
`du=2dx" "v=-e^{-x}`

Let us evaluate the right-hand side.

By Integration by Parts 3,

`(RHS)=-(1+x^2)e^{-x}+int2xe^{-x}dx`

by Integration by Parts 4,

`=-(1+x^2)-2xe^{-x}+int 2e^{-x} dx`

`=-(x^2+2x+1)e^{-x}-2e^{-x}+C`

`=-(x^2+2x+3)e^{-x}+C`

By setting `(LHS)=(RHS)`,

`e^{-y}/2(siny-cosy)=-(x^2+2x+3)e^{-x}+C`

Since `y(0)=0`, we have

`Rightarrow 1/2(0-1)=-3+C Rightarrow C=5/2`

Hence, the solution is implicitly expressed as

`e^{-y}/2(siny-cosy)=-(x^2+2x+3)e^{-x}+5/2`.

I hope that this was helpful.