Question

Dy/dx = root (y-x) ? )

Answer 1

`(sqrt(y-x)-1)e^sqrt(y-x)=sqrt(ce^x)`.

Explanation:

Let, `sqrt(y-x)=v`.

`:. y-x=v^2`.

Diff.ing w.r.t. `x, dy/dx-1=2v*(dv)/dx, or, dy/dx=2v*(dv)/dx+1`.

Sub.ing in the given diff. eqn., we get,

`2v*(dv)/dx+1=sqrt(v^2)=v, i.e.,`

`2v*(dv)/dx=v-1, or,` separating the variables,

`2{v/(v-1)}dv=dx`.

`:. 2{{(v-1)+1}/(v-1)}dv=dx`.

`:. 2{(v-1)/(v-1)+1/(v-1)}dv=dx`.

`:. 2{1+1/(v-1)}dv=dx`.

Integrating, `2int{1+1/(v-1)}dv=intdx+lnc`.

`:.2{v+ln(v-1)}=x+lnc`.

`:. v+ln(v-1)=1/2(lnc+x)`.

`:. lne^v+ln(v-1)=1/2(lnc+lne^x)`.

`:. ln{(v-1)e^v}=1/2{ln(ce^x)}=lnsqrt(ce^x)`.

`:. (v-1)e^v=sqrt(ce^x)`.

Reverting from `v` to `sqrt(y-x)`, we get the Gen. Soln.,

`(sqrt(y-x)-1)e^sqrt(y-x)=sqrt(ce^x)`.