What is the solution for (1+x)y^'+2y=(sin(x))/(1+x) ?

The solution for (1+x)y^'+2y=(sin(x))/(1+x) is y=-(cos(x))/(1+2x+x^2)+(c_{1})/(1+2x+x^2)

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(1+x)y^'+2y=(sin(x))/(1+x)

(1 + x) y^{^{'}} + 2 y = \frac{sin ( x )}{1 + x}

y = - \frac{cos ( x )}{1 + 2 x + x ^{2}} + \frac{c _{1}}{1 + 2 x + x ^{2}}

Solution steps

(1 + x) y^{'} (x) + 2 y = \frac{sin ( x )}{1 + x}

Solve linear ODE:

y = - \frac{cos ( x )}{1 + 2 x + x ^{2}} + \frac{c _{1}}{1 + 2 x + x ^{2}}

y = - \frac{cos ( x )}{1 + 2 x + x ^{2}} + \frac{c _{1}}{1 + 2 x + x ^{2}}

\int sin (5 θ) sin (θ) d θ

\frac{\partial}{\partial x} (9 x e^{4 x y})

\frac{\partial}{\partial x} (\frac{x + y}{2})

\frac{3}{( s + 1 ) ^{2}}

n = 1 \sum \infty \frac{4 ^{n}}{2 ^{n}}

What is the solution for (1+x)y^'+2y=(sin(x))/(1+x) ?
The solution for (1+x)y^'+2y=(sin(x))/(1+x) is y=-(cos(x))/(1+2x+x^2)+(c_{1})/(1+2x+x^2)