What is the solution for (1+x^2)y^{''}+y^'x+ax=0 ?

The solution for (1+x^2)y^{''}+y^'x+ax=0 is y=c_{1}ln(sqrt(x^2+1)+x)-ax+c_{2}

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(1+x^2)y^{''}+y^'x+ax=0

(1 + x^{2}) y^{^{''}} + y^{^{'}} x + a x = 0

y = c_{1} ln (x^{2} + 1 + x) - a x + c_{2}

Solution steps

(1 + x^{2}) y^{''} (x) + y^{'} (x) x + a x = 0

Solve using substitution:

y = c_{1} ln (x^{2} + 1 + x) - a x + c_{2}

y = c_{1} ln (x^{2} + 1 + x) - a x + c_{2}

f (t) = 3 cos (t)

\int \frac{sec ( x )}{2 sec ( x ) + 1} d x

x^{2} + y^{2} = 100, (- 6, 8)

\int - x cos (\frac{n x π}{l}) d x

\frac{d}{d x} (4 x^{3} - 3)

What is the solution for (1+x^2)y^{''}+y^'x+ax=0 ?
The solution for (1+x^2)y^{''}+y^'x+ax=0 is y=c_{1}ln(sqrt(x^2+1)+x)-ax+c_{2}