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Apply Rungee Kutta Method of Fourth Order to Find an Approximate Value of Y When X=0.4 Given that D Y D X = Y βˆ’ X Y + X , Y = 1 π’š=𝟏 π’˜π’‰π’†π’ 𝒙=𝟎 Taking H=0.2. - Applied Mathematics 2

Apply Rungee Kutta method of fourth order to find an approximate Value of y when x=0.4 given that `dy/dx=(y-x)/(y+x),y=1` π’š=𝟏 π’˜π’‰π’†π’ 𝒙=𝟎 Taking h=0.2. 

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Solution

(1) `dy/dx=(y-x)/(y+x)     x_0=y_0=1, h=0.2` 

𝒇(𝒙,π’š)=` (y-x)/(y+x)` 

π’ŒπŸ=𝒉.𝒇(π’™πŸŽ,π’šπŸŽ)=𝟎.πŸπ’‡(𝟎,𝟏)=𝟎.𝟐  

`k_2=h.f(x_0+h/2,y_0+k_1/2)=0.2 f (0.1,1.1)=0.1661` 

`k_3=h.f (x_0+h/2,y_0+k_2/2)=0.2.f(0.1.1.0833)=0.1661`

`k_4=h.f (x_0+h,y_0+k_3)=0.2f(0.2,.1.1661)=0.1414` 

`k= (k_1+2k_2+2k_3+k_4)/6=(0.2+2(0.1666)+2(0.1661)+0.1414)/6=0.1678` 

∴` y(0.2)=y_0+k=1+0.1678=1.1678`

(II) `x_1=0.2,y_2=1.1678,h=0.2` 

`k_5=h.f(x_1,y_1)=0.2f(0.2,1.1678)=0.1415`

`k_6=h.f(x_1+h/2,y_1 ++k_5/2)=0.2 f(0.3,1.2285)=0.1220`

`k_7=h.f(x_1+h/2,y_1 ++k_6/2)=0.2f(0.3,1.2285)=0.1214` 

`k_8=h.f(x_1+h,y_1+k_7)=0.2f(0.4,1.2892)=0.10`

`k* =( k_5+2k_6+2k_7+k_8)/6 = (0.1415+2(0.1215)+2(0.1215)+0.1052)/6=0.1222` 

π’š(𝟎.πŸ’)=π’šπŸ+π’Œ∗=𝟏.πŸπŸ”πŸ•πŸ–+𝟎.𝟏𝟐𝟐𝟐=𝟏.πŸπŸ—πŸŽ 

 

Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
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