# 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 β X Y + X , Y = 1 π=π ππππ π=π 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=π ππππ π=π Taking h=0.2.

#### Solution

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

π(π,π)= (y-x)/(y+x)

ππ=π.π(ππ,ππ)=π.ππ(π,π)=π.π

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(π.π)=ππ+π∗=π.ππππ+π.ππππ=π.πππ

Concept: Linear Differential Equation with Constant Coefficientβ Complementary Function
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