1. Verify that $y^2=e^{2x}+c$ is a solution of the differential equation $yy'=e^{2x}$.

Verify:Let $F(x,y)=y^2-e^{2x}-c$,then according to the implicit function theorem,$\frac{dy}{dx}=\frac{2e^{2x}}{2y}=\frac{e^{2x}}{y}$.So $yy'=e^{2x}$

2.$y=c_1\sin 2x+c_2\cos 2x$ is a solution of $y''+4y=0$.

Proof:$y''=-4c_1\sin 2x-4c_2\cos 2x$.So $y''+4y=0$.

3.$y=c_1e^{2x}+c_2e^{-2x}$ is a solution of $y''-4y=0$.

Proof:$y''=4c_1e^{2x}+4c_2e^{-2x}$.So $y''=4y$.

4.$y=c_1\sinh 2x +c_2\cosh 2x$ is a solution of $y''-4y=0$.

Proof:$\sinh 2x=\frac{e^{2x}-e^{-2x}}{2}$.$\cosh 2x=\frac{e^{2x}+e^{-2x}}{2}$.According to 3, both $\sinh 2x$ and $\cosh 2x$ is a solution of $y''-4y=0$,so $c_1\sinh 2x+c_2\cosh 2x$ is a solution of $y''-4y=0$.

5.$y=\sin^{-1}xy$ is a solution of $xy'+y=y' \sqrt{1-x^2y^2}$.

Proof:Let $F(x,y)=y-\sin^{-1}xy$.$\frac{dy}{dx}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}(x,y)$.And $\frac{\partial F}{\partial

x}(x,y)=\frac{-y}{\sqrt{1-x^2y^2}}$.$\frac{\partial F}{\partial y}(x,y)=1-\frac{x}{\sqrt{1-x^2y^2}}$.So $\frac{dy}{dx}=\frac{y}{\sqrt{1-x^2y^2}-x}$.So $xy'+y=y' \sqrt{1-x^2y^2}$.

6.$y=x\tan x$ is a solution of $xy'=y+x^2+y^2$.

Proof:$y'=\tan x+\frac{x}{\cos^2x}=\tan x+x(1+\tan^2x)$.So $xy'=y+x^2+y^2$.

 

7.$x^2=2y^2\log y$ is a solution of $y'=\frac{xy}{x^2+y^2}$.

Proof:$\frac{dy}{dx}=\frac{x}{2y\log y+y}$.So $\frac{dy}{dx}=\frac{xy}{x^2+y^2}$.

8.$y^2=x^2-cx$ is a solution of $2xyy'=x^2+y^2$.

Proof:$\frac{dy}{dx}=\frac{2x-c}{2y}$.So $2xyy'=2x^2-cx=x^2+y^2$.

9.$y=ce^{\frac{y}{x}}$ is a solution of $y'=\frac{y^2}{xy-x^2}$.

Proof:$\frac{dy}{dx}=-\frac{ce^{\frac{y}{x}}(\frac{-y}{x^2})}{1-ce^{\frac{y}{x}}\frac{1}{x}}=\frac{y^2}{xy-x^2}$.

10.$y+\sin y=x$ is a solution of $(y\cos y-\sin y+x)y'=y$.

Proof:$\frac{dy}{dx}=\frac{1}{1+\cos y}$.So $(y\cos y-\sin y+x)y'=y$.

11.$x+y=\tan^{-1}y$ is a solution of $1+y^2+y^2y'=0$.

Proof:$\tan (x+y)=y$.So $\frac{dy}{dx}=\frac{-1}{\sin^2(x+y)}$.So it is easy to verify that $1+y^2+y^2y'=0$.