Question

2dy/dx-2y=x⁵sin2x-x³+4x⁴

Answer 1

To solve linear equation :

$2\frac{dy}{dx}-2y=x^5\sin{2x}-x^3+4x^4$

Lets divide both sides of equation by 2 and multiply both sides by $f(x)=e^{\int-1dx'}=e^{-x}$

$e^{-x}\frac{dy}{dx}-e^{-x}y=\frac{e^{-x}}{2}(x^5\sin{2x}-x^3+4x^4)$

Lets substitute $-e^{-x}=\frac{d}{dx}(e^{-x})$

$e^{-x}\frac{dy(x)}{dx}+\frac{d}{dx}(e^{-x})y(x)=\frac{1}{2}e^{-x}(x^5\sin{2x}-x^3+4x^4)$

Apply the reverse product rule $f\frac{dg}{dx}+g\frac{df}{dx}=\frac{d}{dx}(fg)$ to left-hand side:

$\frac{d}{dx}(e^{-x}y(x))= \frac{1}{2}e^{-x}(x^5\sin{2x}-x^3+4x^4)$

Integrate both sides with respect to x:

$\int\frac{d}{dx}(e^{-x}y(x))dx= \int\frac{1}{2}e^{-x}(x^5\sin{2x}-x^3+4x^4)dx$

Evaluate the integrals:

$e^{-x}y(x)= \frac{1}{6250}e^{-x}(-3125(4x^4+15x^3+45x^2+90x+90)-2\cos(2x)(625x^5+1250x^4-500x^3-3600x^2-2280x+528)+(-625x^5+1875x^4+5500x^3+2100x^2-4920x-2808)\sin(2x))+c_1$

by dividing both sides by $e^{-x}$

$y(x)= \frac{e^x}{6250}e^{-x}(-3125(4x^4+15x^3+45x^2+90x+90)-2\cos(2x)(625x^5+1250x^4-500x^3-3600x^2-2280x+528)+(-625x^5+1875x^4+5500x^3+2100x^2-4920x-2808)\sin(2x))+c_1$