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

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

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

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

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

Solution steps

(1 + x^{2}) y^{'} (x) + 4 x y = x

Solve separable ODE:

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

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

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