Question

Solve by method of variation of parameters 

`(d^2y)/dx^2+3 dy/dx+2y=e^(e"^x)` 

Answer 1

A.E: 𝐷²+3𝐷+2=0  

Solving the equation, we get 

∴` D=-1,-2`  

∴` C.F=C_1 e^-x +C_2 e^-(2x)` 

∴` y_1=e^-x  y_2 = e^(-2x)`  

∴ `y'1=-e^-x`  `  y'2=-2e^(-2x)` 

∴ W=`|[y_1,y_2],[y'_1,y'_2]| = |[e_-x,e^(-2x)],[-e^-x,-2e^(-2x)]|`  

=`-2e^(-3x)+e^(-3x)` 

=`-e^(-3x)` 

`x=e^(e^x)` 

∴ `u=- int (y^2x)/W dx` 

=`-int (e^-2x e^e"^x)/-e^-3x  dx ` 

= `-int ^(e^e"^x).e^x  dx` 

Put `e^x=t`

`e^x dx=dt` 

∴` int e^t  dt=e^t+c.`

∴` W=e^(e^x)+c` 

`v= int (y_1 X)/w dx` 

`v= int (e^-x e^(e^x))/(-e-3x)  dx`  

`V= int e^(e^x) e^(2x) dx` 

Putting` e^x=t` 

∴` v= int e^t.  t dt= te^t-e^t` 

∴ `v= e^x e^(e^x)-e^(e^x)` 

∴` P.I=uy_1+vy_2= e^(e^x) . e^-x- (e^x e^(e^x)-e^(e^x))e^(-2x)` 

=`e^-2x. e^(e^x)` 

∴ The complete solution is,

y=C.f+P.I 

`y=c_1 e^-x+C_2e^- 2x +e^-2x.e^e^x`

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