What is the solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 ?

The solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 is y= 1/3+(16)/(3(x^2+4)^{3/2)}

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y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1

y^{^{'}} + \frac{3 x}{x ^{2} + 4} y = \frac{x}{x ^{2} + 4}, y (0) = 1

y = \frac{1}{3} + \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}}

Solution steps

y^{'} (x) + \frac{3 x}{x ^{2} + 4} y = \frac{x}{x ^{2} + 4}

Solve linear ODE:

y = \frac{1}{3} + \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}}

y = \frac{1}{3} + \frac{16}{3 ( x ^{2} + 4 ) ^{\frac{3}{2}}}

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\frac{d}{d x} (\frac{36}{x ^{2} + 2})

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What is the solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 ?
The solution for y^'+(3x)/(x^2+4)y= x/(x^2+4),y(0)=1 is y= 1/3+(16)/(3(x^2+4)^{3/2)}