Question

Q:Solve the DE\[(1+x^2)y^{\prime\prime} = 1\]

A:\[y=x\arctan x-\frac 12\ln\left(1+x^2\right)+cx+d\]

Answer 1

\[(1+x^2)y^{\prime\prime} = 1\]\[y^{\prime\prime}=\frac{1}{1+x^2}\]\[\text{let } y^\prime=p\]\[y^{\prime\prime}=p^\prime\]\[p^\prime=\frac{1}{1+x^2}\]\[\int\text dy=\int\frac{\text dx}{1+x^2}\]\[y=\arctan x+c\]

Answer 2

how do i get the other terms in the answer?

Answer 3

y is not the arctan(x)

Answer 4

y' is

Answer 5

integrate again...by parts

Answer 6

i dont understand @Zarkon

Answer 7

\[y^{\prime\prime}=\frac{1}{1+x^2}\]

\[\int y^{\prime\prime}dx=\int\frac{1}{1+x^2}dx\]

\[y'=\arctan(x)+c\]


\[\int y'dx=\int(\arctan(x)+c)dx\]

\[y=\int(\arctan(x)+c)dx=\cdots\]

Answer 8

the 'answer' you have in the original post is not correct

Answer 9

ops, ill fix that sorry

Answer 10

\[y=x\cdot \arctan(x)-\frac 12\ln\left(1+x^2\right)+cx+d\]

Answer 11

\[\int\text dp=\int\frac{\text dx}{1+x^2}\]\[p=\arctan x+c\]\[y^\prime=\arctan x+c\]\[y=\int\left(\arctan x+c\right)\text dx\]

Answer 12

\[u=\arctan x\qquad\qquad v^\prime=1\]\[u^\prime=\frac{1}{1+x^2}\qquad\qquad v=x\]\[y=x\arctan x-\int \frac{x}{1+x^2}\cdot\text dx+\int c\cdot\text dx\]\[y=x\arctan x-\frac 12 \ln|1+x^2|+cx+d\]

Answer 13

you can drop the absolute value if you want since \(1+x^2>0\), \(\forall x\in\mathbb{R}\)

Answer 14

ah that is good to know.
\[y=x\arctan x-\frac 12 \ln\left(1+x^2\right)+cx+d \qquad\checkmark\]