Question

Решить дифференциальное уравнение

Answer 1

$displaystyle y``=frac{1}{1+x^2}\\int {y``} , dy=int{frac{1}{1+x^2}} , dx \\y`=arctgx+C1\\y`(0)=0; arctg0+C1=0; C1=0\\int {y`} , dy=int {arctgx} dx\\U=arctgx; dU=frac{1}{x^2+1}dx\\dV=dx; V=x\\ y=int{U(x)*dV} ,=U(x)*V(x)-int {V(x)dU}\\y=arctgx*x-int {frac{x}{1+x^2}dx }=x*arctgx-frac{ln(x^2+1)}{2}+C2\\y(0)=0; 0*arctg 0-frac{ln1}{2}+C2=0; C2=0\\y=x*arctgx-frac{ln(x^2+1)}{2}$

$displaystyle y(1)=1*arctg 1-frac{ln(2)}{2} approxfrac{pi}{4}-frac{0.69}{2}=0.785-0.345=0.44$