Question

For each of the differential equations, find the general solution:
1. $\frac{dy}{dx}+2y=\sin x$

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

3. $\frac{dy}{dx}+\frac{y}{x}=x^2$
4. $\frac{dy}{dx}+\left(\sec x\right)y=\tan x$ ...$(0\le x<\frac{\pi }{2})$
5. ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$ ...$(0\le x<\frac{\pi }{2})$
6. $x\frac{dy}{dx}+2y=x^2\log x$

7. $x\log x\frac{dy}{dx}+y=\frac{2}{x}\log x$
8. $(1+x^2)dy+2xydx=\cot xdx$ ...$(x\ne 0)$
9. $x\frac{dy}{dx}+y-x+xy\cot x=0$ ...$(x\ne 0)$

10. $(x+y)\frac{dx}{dy}=1$
11. $ydx+(x-y^2)dy=0$

12. $(x+3y^2)\frac{dy}{dx}=y$ ...$(y>0)$

Answer 1

(1) $\frac{dy}{dx}+2y=\sin \theta \frac{dy}{dx}+Py=QI.F=e^{\int Pdx}=e^{2x}G.S=y(I.F)=\int Q(I.F)dx{\Rightarrow }^{ax}\sin bx=\frac{a^{ax}}{a^2+b^2}x\Rightarrow (a\sin bx-b\cos bx)+Cy\left(e^{2x}\right)=\int \sin x{e}^{2x}dxy\left(e^{2x}\right)=\frac{e^{2x}}{5}(2\sin x-\cos x)+C2.\frac{dy}{dx}+3y=e^{-2x}\frac{dy}{dx}+Py=QI.F=e^{\int Pdx}=e^{2x}=e^{3x}G.S=y(I.F)=\int Q(I.F)dx=y\left(e^{3x}\right)=\int e^{-2x}{e}^{3x}dx=y\left(e^{3x}\right)=\int e^{x}dx=y\left(e^{3x}\right)=e^x+C3.\frac{dy}{dx}+\frac{y}{x}=x^2\frac{dy}{dx}+Py=QI.F=e^{\int Pdx}=e^{\int \frac{1}{x}dx}=e^{\log x}=xG.S=y(I.F)=\int Q(I.F)dx=y\left(x\right)=\int x^2\left(x\right)dx=xy=\frac{x^3}{3}+C=x^3-3xy+k=04.\frac{dy}{dx}+\left(\sec x\right)y=\tan x\frac{dy}{dx}+Py=QI.F=e^{\int \sec xdx}=e^{\log |\sec x+\tan x|}=\sec x+\tan xG.S=y(I.F)=\int Q(I.F)dx\Rightarrow y(\sec x+\tan x)=\int \sec x\tan xdx+\int {\tan }^{2}dx=\sec x+\int ({\sec }^{2}x-1)dx\Rightarrow y(\sec x+\tan x)=\sec x+\tan x-x+C5.{\cos }^2\frac{dy}{dx}+y=\tan x\frac{dy}{dx}+{\sec }^{2}x\left(y\right)=\frac{\tan x}{{\cos }^{2}x}=\frac{\sin x}{{\cos }^{3}x}I.F=e^{\int pdx}=e^{\int {\sec }^{2}xdx}=e^{\tan x}G.S=y(I.F)=\int Q(I.F)dx=y\left(e^{\tan x}\right)=\int \tan x{e}^{\tan x}{\sec }^{2}dx=y\left(e^{\tan x}\right)=\int \tan x{\sec }^{2}x{e}^{\tan x}dx+Clet\tan x=t{\sec }^{2}xdx=dty{e}^t=\int t{e}^{t}dt+C\Rightarrow y{e}^{\tan x}=(\tan x-1)e^{\tan x}+C$

$6.x\frac{dy}{dx}+2y=x^2\log x\frac{dy}{dx}+\frac{2}{x}\left(y\right)=x\log x\frac{dy}{dx}+Py=QI.F=e^{\int Pdx}=e^{\int \frac{2}{x}dx}=e^{2\log x}=x^{2}G.Sy(I.F)=\int Q(I.F)dx=y\left(x^2\right)=\int x^3\log xdxy\left(x^2\right)=\log x\left(\frac{x^4}{4}\right)-\int \frac{1}{x}\left(\frac{x^4}{4}\right)dx{x}^{2}y=\frac{x^4\log x}{4}-\frac{1}{16}{x}^4+C7.x\log x+y=\frac{2}{x}\log x\frac{dy}{dx}+\frac{1}{x\log x}\left(y\right)=\frac{2}{x^2}I.F=e^{\int \frac{1}{x\log x}dx}=\log xG.Sy(I.F)=\int Q(I.F)dx=y\left(\log x\right)=\int \frac{2}{x^2}\log xdx=\int \frac{1}{x^2}\log xdx=\log x(-\frac{1}{x})-\int \frac{1}{x}(-\frac{1}{x})dx=-\log x\left(\frac{1}{x}\right)+\int \frac{1}{x^2}dx=-\frac{1}{x}\log x-\frac{1}{x}Csolution\Rightarrow y\left(\log x\right)=-\frac{1}{x}\log x-\frac{1}{x}+C8.(1+x^2)dy+2xydx=\cot xdx(1+x^2)\frac{dy}{dx}=\cot x-2xy\frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\frac{\cot x}{1+x^2}I.F=e^{\int Pdx}=e^{\int \frac{2x}{1+x^2}dx}=1+x^{2}G.Sy(I.F)=\int Q(I.F)dxy(1+x^2)=\int \frac{\cot x}{1+x^2}\times 1+x^{2}dxy(1+x^2)=\log \left|\sin x\right|+C9.x\frac{dy}{dx}+y-x+xy\cot x=0\frac{dy}{dx}+\frac{1}{x}\left(y\right)-1+y\cot x=0\frac{dy}{dx}+[\frac{1}{x}+\cot x]y=1I.F=e^{\int Pdx}=e^{\int \frac{1}{x}+\cot xdx}=e^{\log x+\log \lfloor \sin x\rfloor }G.Sy(I.F)=\int Q(I.F)dx=y\left(x\sin x\right)=\int x\sin xdx=y\left(x\sin x\right)=-x\cos x+\sin x+C10.(x+y)\frac{dy}{dx}=1\frac{dy}{dx}-x=y\frac{dy}{dx}+Px=QI.F=e^{\int Pdy}=e^{\int -1dy}=e^{-y}G.S=x\left(e^{-y}\right)=\int y{e}^{-y}dyx\left(e^{-y}\right)=-y{e}^{-y}-e^{-y}+C$

$11.ydx+(x-y^2)dy=0ydx+(y^2-x)dy\frac{dx}{dy}=y-\frac{x}{y}\frac{dx}{dy}+\frac{x}{y}=yI.F=e^{\int Pdy}=e^{\int \frac{1}{y}dy}=yG.S=x\left(y\right)=\int y\left(y\right)dy=x\left(y\right)=\frac{y^3}{3}+C12.(x+{3y}^2)\frac{dy}{dx}=y\frac{dx}{dy}=\frac{x}{y}+3y\frac{dx}{dy}-\frac{x}{y}=3yI.F=e^{\int Pdy}=e^{\int -\frac{1}{y}dy}=\frac{1}{y}G.S=x\left(\frac{1}{y}\right)=\int 3y\left(\frac{1}{y}\right)dy\frac{x}{y}=3y+C$