Question

(1+x^3)dy/dx+6x^2y=e^x​

Answer 1

Step-by-step explanation:

We have,

$(1 + {x}^{3} ) \frac{dy}{dx} + 6 {x}^{2} y = {e}^{x}$

$\implies \frac{dy}{dx} + \frac{6 {x}^{2} }{ {x}^{3} + 1 }. y = \frac{ { e }^{x} }{ {x}^{3} + 1 }$

Now,

$I.F.= {e}^{2 \int \frac{3 {x}^{2} }{ {x}^{3} + 1 } dx} = {e}^{2 ln( {x}^{3} + 1) } = ( {x}^{3} + 1)^{2}$

$\implies \: y( {x}^{3} + 1)^{2} = \int \frac{ {e}^{x} }{ {x}^{3} + 1} .( {x}^{3} + 1)^{2} dx$

$\implies \: y( {x}^{3} + 1)^{2} = \int {e}^{x} ( {x}^{3} + 1)dx$

$\implies \: y( {x}^{3} + 1)^{2} = {e}^{x}( {x}^{3} + 1) - {e}^{x} (3 {x}^{2} ) - {e}^{x}.(6x) - {e}^{x} (6) + c$

$\implies \: y( {x}^{3} + 1)^{2} = {e}^{x} ( {x}^{3} - 3 {x}^{2} - 6x - 5) + c$