# Solve by Variation of Parameter Method D 2 Y D X 2 + 3 D Y D X + 2 Y = E E X . - Applied Mathematics 2

Sum

Solve by variation of parameter method (d^2y)/(dx^2)+3(dy)/(dx)+2y=e^(e^x).

#### Solution

(d^2y)/(dx^2)+3(dy)/(dx)+2y=e^(e^x).

Put D=d/(dx)   thereforeD^2y+3Dy+2y=0

For complementary solution,
f(D)=0
thereforeD^2+3D+2=0

𝑫= −𝟏 ,−𝟐

therefore y_c=c_1e^(x)+c_2e^(-2x)

Particular integral is given by ,

y_p=y_1p_1+y_2p_2

wherep_1=int(-y_2x)/w dx

p_1=int(y_1x)/w dx

w=|(y_1,y_2),(y'_1,y'_2)|

therefore w=|(e^(-x),e^(-2x)),(-e^(-x),-2e^(-2x))|=-e^(-3x)

p_1=int(e^(-2x)e^(e^x))/e^(-3x) dx=inte^(e^x)e^xdx=inte^t dt=e^(e^x) ........{"Put"  e^x=t=>e^xdx=dt}

p_2=int(e^(-2x))/e^(-3x)e^(e^x) dx=inte^(e^x)e^(2x) dx=int t e^t dt=e^x e^(e^x)-e^(e^x)

therefore y_p=e^x e^(e^x)-(e^x e^(e^x)-e^(e^x))e^(-2x)=e^(-2x) e^(e^x)

The general solution of given differential eqn is given by ,

y_g=y_c+y_p=c_1e^(-x)+c_2e^(-2x)+e^(-2x)e^(e^x)

Concept: Method of Variation of Parameters
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