Solve y’=y-x2
, y(0)=1, by Picord’s method upto the third approximation . Hence find
the value of y(0,1), y(0,2)
y(t,0)=1y(t,1)=y(0)+∫0t(y(s,0)−s2)ds=1−t33y(t,2)=y(0)+∫0t(y(s,1)−s2)ds=1+t−t412−t33y(t,3)=y(0)+∫0t(y(s,2)−s2)ds=1+t+t22−t560−t412y(0,1)=y(0,2)=1y\left( t,0 \right) =1\\y\left( t,1 \right) =y\left( 0 \right) +\int_0^t{\left( y\left( s,0 \right) -s^2 \right) ds}=1-\frac{t^3}{3}\\y\left( t,2 \right) =y\left( 0 \right) +\int_0^t{\left( y\left( s,1 \right) -s^2 \right) ds}=1+t-\frac{t^4}{12}-\frac{t^3}{3}\\y\left( t,3 \right) =y\left( 0 \right) +\int_0^t{\left( y\left( s,2 \right) -s^2 \right) ds}=1+t+\frac{t^2}{2}-\frac{t^5}{60}-\frac{t^4}{12}\\y\left( 0,1 \right) =y\left( 0,2 \right) =1y(t,0)=1y(t,1)=y(0)+∫0t(y(s,0)−s2)ds=1−3t3y(t,2)=y(0)+∫0t(y(s,1)−s2)ds=1+t−12t4−3t3y(t,3)=y(0)+∫0t(y(s,2)−s2)ds=1+t+2t2−60t5−12t4y(0,1)=y(0,2)=1
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