Question

Решить дифференциальное уравнение

Answer 1

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

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