Answer:
Step-by-step explanation:
Consider the initial given value problem
[tex]y'=-x-y,y(1)=1[/tex]
Use Euler's Method
[tex]\Delta =0.1[/tex]
Then, we have
[tex]y(t_1=1.1)=y(t_0+\Delta)\\\\=y_0+\Delta y'(t_0)\\=1+(0.1)(-1-1)\\=1-0.2\\=0.8\\=y_1[/tex]
[tex]y(1.2)=y_1+\Delta y'(t_1,y_1)\\=0.8+0.1(-1.1-0.80\\=0.61\\=y_2\\\\y(1.3)=y_2+\Delta y'(t_2,y_2)\\=(0.61)+(0.1)(-1.2-0.61)\\=0.429\\=y_3\\\\y(1.4)=(0.429)+0.1(-1.3-0.429)\\=0.2561[/tex]
Now estimate y when x =2.4 for the solution curve satisfy y(1) = 0
Then,
[tex]y(1.1)=y_0+ \Delta y'(t_0)=0+(0.1)(-1-0)=-0.1[/tex]
[tex]y(1.2)=-0.1+0.1(-1.1+0.1)=-0.2\\y(1.3)=-0.2+0.1(-1.2+0.2)=-0.3\\y(1.4)=-0.3+0.1(-1.3+0.3)=-0.4\\y(1.5)=-0.4+0.1(-1.4+0.4)=-0.5\\y(1.6)=-0.5+0.1(-1.5+0.5)=-0.6\\y(1.7)=-0.6+0.1(-1.6+0.6)=-0.7\\y(1.8)=-0.7+0.1(-1.7+0.7)=-0.8\\y(1.9)=-0.8+0.1(-1.8+0.8)=-0.9\\y(2.0)=-0.9+0.1(-1.9+0.9)=-1.0\\y(2.1)=-1.0+0.1(-2.0+1)=-1.1\\y(2.2)=-1.1+0.1(-2.1+1.1)=-1.2\\y(2.3)=-1.2+0.1(-2.2+1.2)=-1.3\\y(2.4)=-1.3+0.1(-2.3+1.3)=-1.4[/tex]
Hence,Euler's approximation y(2.4) = -1.4
