What is the particular solution of the differential equation? : dy/dx + 1/x = e^y/x dydx+1x=eyx with y(1)=ln2y(1)=ln2
1 Answer
y = ln(2/(2-x)) y=ln(22−x)
Explanation:
We have:
dy/dx + 1/x = e^y/x dydx+1x=eyx withy(1)=ln2y(1)=ln2 ..... [A]
This is a First Order non-linear Differential Equation. The
e^y=t => y=lnt => dy/dt = 1/t ey=t⇒y=lnt⇒dydt=1t
As we are changing variable,m we change the initial conditions accordingly:
When
y=ln2 => t=2y=ln2⇒t=2
Substituting into the original differential equation [A], we get:
:. dy/dt dt/dx + 1/x = e^y/x
:. 1/t dt/dx + 1/x = t/x
:. 1/t dt/dx = (t-1)/x
:. 1/(t(t-1)) dt/dx = 1/x
This has reduced the DE [A] to a First Order Separable equation, so we can "seperate the variables" to get:
int \ 1/(t(t-1)) \ dt = int \ 1/x \ dx ..... [B]
The LHS integral is trivial, and for the RHS integral we must decompose the integrand into partial fractions:
1/(t(t-1)) -= A/t + B/(t-1)
" " = (A(t-1) + Bt)/ ( t(t-1) )
Leading to the identity:
1 -= A(t-1) + Bt
We can find the coefficients
Put
t=0 => 1=A(-1)+0 => A=-1
Putt=1 => 1=0+B \ \ \ \ \ \ \ \ \ \=> B=1
Using this result, we can now integrate [B] to get the general solution:
int \ 1/(t-1) -1/t \ dt = int \ 1/x \ dx
:. ln|t-1| - ln|t| = ln |x| + C
Using the initial condition
:. ln1 - ln2 = ln 1+ C => C=-ln2
And so the particular solution:
ln|t-1| - ln|t| = ln |x| -ln2
:. ln|(t-1)/t| = ln |x/2|
:. (t-1)/t = x/2
:. 2t-2 = tx
:. t(2-x)=2
:. t = 2/(2-x)
Restoring the substitution, we have:
y = lnt
\ \ = ln(2/(2-x))
