Question

Find the general solution of each of the following

i) (2xsiny+y³e^x)dx +(x²cosy+3y²e^x)dy=0

ii) (ysec²x+secxtanx)dx + (tanx+2y)dy=0

iii)(ye^x+2e^x+y²)dx+(e^x+2xy)dy=0 y(0)=6

iv)(2xcosy+3x²y)dx+(x³-x²siny-y)dy=0 y(1)=3

Answer 1

i)

$$(2x\sin y+y^3e^x)dx +(x^2\cos y+3y^2e^x)dy=0$$

$$P=2x\sin y+y^3e^x, \dfrac{\partial P}{\partial y}=2x\cos y+3y^2e^x$$

$$Q=x^2\cos y+3y^2e^x, \dfrac{\partial Q}{\partial x}=2x\cos y+3y^2e^x$$

$$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$$

$$\dfrac{\partial u}{\partial x}=P$$

$$\dfrac{\partial u}{\partial y}=Q$$

$$u(x,y)=\int P dx+\varphi(y)$$

$$=\int (2x\sin y+y^3e^x)dx+\varphi(y)$$

$$=x^2\sin y+y^3e^x+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=x^2\cos y+3y^2e^x+\varphi'(y)$$

$$=x^2\cos y+3y^2e^x=Q$$

Then $\varphi'(y)=0$

$$\varphi(y)=C_1$$

The general solution of the equation is defined by the following implicit expression:

$$x^2\sin y+y^3e^x=C,$$

where $C$ is an arbitrary real number.

ii)

$$(y\sec^2x+\sec x\tan x)dx + (\tan x+2y)dy=0$$

$$P=y\sec^2x+\sec x\tan x, \dfrac{\partial P}{\partial y}=\sec^2x$$

$$Q=\tan x+2y, \dfrac{\partial Q}{\partial x}=\sec^2x$$

$$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$$

$$\dfrac{\partial u}{\partial x}=P$$

$$\dfrac{\partial u}{\partial y}=Q$$

$$u(x,y)=\int P dx+\varphi(y)$$

$$=\int (y\sec^2x+\sec x\tan x)dx+\varphi(y)$$

$$=y\tan x+\sec x+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=\tan x+\varphi'(y)$$

$$=\tan x+2y=Q$$

Then $\varphi'(y)=2y$

$$\varphi(y)=y^2+C_1$$

The general solution of the equation is defined by the following implicit expression:

$$y\tan x+\sec x+y^2=C,$$

where $C$ is an arbitrary real number.

iii)

$$(ye^x+2e^x+y^2)dx+(e^x+2xy)dy=0$$

$$P=ye^x+2e^x+y^2, \dfrac{\partial P}{\partial y}=e^x+2y$$

$$Q=e^x+2xy, \dfrac{\partial Q}{\partial x}=e^x+2y$$

$$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$$

$$\dfrac{\partial u}{\partial x}=P$$

$$\dfrac{\partial u}{\partial y}=Q$$

$$u(x,y)=\int P dx+\varphi(y)$$

$$=\int (ye^x+2e^x+y^2)dx+\varphi(y)$$

$$=ye^x+2e^x+xy^2+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=e^x+2xy+\varphi'(y)$$

$$=e^x+2xy=Q$$

Then $\varphi'(y)=0$

$$\varphi(y)=C_1$$

The general solution of the equation is defined by the following implicit expression:

$$ye^x+2e^x+xy^2=C,$$

where $C$ is an arbitrary real number.

$y(0)=6$

$$6e^0+2e^0+0(6)^2=C$$

$$C=8$$

The solution of the given equation is

$$ye^x+2e^x+xy^2=8$$

iv)

$$(2x\cos y+3x^2y)dx+(x^3-x^2\sin y-y)dy=0$$

$$P=2x\cos y+3x^2y, \dfrac{\partial P}{\partial y}=-2x\sin y+3x^2$$

$$Q=x^3-x^2\sin y-y, \dfrac{\partial Q}{\partial x}=3x^2-2x\sin y$$

$$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$$

$$\dfrac{\partial u}{\partial x}=P$$

$$\dfrac{\partial u}{\partial y}=Q$$

$$u(x,y)=\int P dx+\varphi(y)$$

$$=\int (2x\cos y+3x^2y)dx+\varphi(y)$$

$$=x^2\cos y+x^3y+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=-x^2\sin y+x^3+\varphi'(y)$$

$$=x^3-x^2\sin y-y=Q$$

Then $\varphi'(y)=-y$

$$\varphi(y)=-\dfrac{1}{2}y^2+C_1$$

The general solution of the equation is defined by the following implicit expression:

$$2x\cos y+3x^2y-\dfrac{1}{2}y^2=C,$$

where $C$ is an arbitrary real number.

$y(1)=3$

$$2(1)\cos 3+3(1)^2(3)-\dfrac{1}{2}(3)^2=C$$

$$C=2\cos3+4.5$$

The solution of the given equation is

$$2x\cos y+3x^2y-\dfrac{1}{2}y^2=2\cos3+4.5$$