Solve the following differential equation
Answers
Answer:
(y
2
e
xy
2
+4x
3
)dx+(2xye
xy
2
−3y
2
)dy=0
y
2
1
dx
dy
+
y
1
=2x
\ -\frac{1}{y^2} \frac{dy}{dx}-\frac{1}{y}=-2x −
y
2
1
dx
dy
−
y
1
=−2x
v(x)=\frac{1}{y(x)}, \ \frac{dv}{dx}=-\frac{1}{y^2(x)}\frac{dy}{dx}v(x)=
y(x)
1
,
dx
dv
=−
y
2
(x)
1
dx
dy
−v=−2x
Multiply by e^{-x}e
−x
both sides of equation:
e^{-x}\frac{dv}{dx} -e^{-x}v=-2xe^{-x}e
−x
dx
dv
−e
−x
v=−2xe
−x
e^{-x}\frac{dv}{dx} +v\frac{d}{dx}(e^{-x})=-2xe^{-x}e
−x
dx
dv
+v
dx
d
(e
−x
)=−2xe
−x
\frac{d}{dx}(ve^{-x})=-2xe^{-x}
dx
d
(ve
−x
)=−2xe
−x
ve^{-x}=\int (-2xe^{-x})dx=2xe^{-x}+2e^{-x}+Cve
−x
=∫(−2xe
−x
)dx=2xe
−x
+2e
−x
+C
Return to y(x):y(x):
\frac{1}{y}e^{-x}= 2xe^{-x}+2e^{-x}+C
y
1
e
−x
=2xe
−x
+2e
−x
+C
y=\frac{e^{-x}}{2xe^{-x}+2e^{-x}+C}=\frac{1}{2x+2+Ce^{x}}y=
2xe
−x
+2e
−x
+C
e
−x
=
2x+2+Ce
x
1
Answer: y(x)=\frac{1}{Ce^x+2(x+1)}y(x)=
Ce
x
+2(x+1)
1
1.2) (y^2e^{xy^2}+4x^3)dx+(2xye^{xy^2}-3y^2)dy=01.2)(y
2
e
xy
2
+4x
3
)dx+(2xye
xy
2
−3y
2
)dy=0
P(x,y)=y^2e^{xy^2}+4x^3, \ \ Q(x,y)=2xye^{xy^2}-3y^2P(x,y)=y
2
e
xy
2
+4x
3
, Q(x,y)=2xye
xy
2
−3y
2
\frac{\partial P(x,y)}{\partial y} =\frac{\partial Q(x,y)}{\partial x}=2ye^{xy^2}+2xy^3e^{xy^2}
∂y
∂P(x,y)
=
∂x
∂Q(x,y)
=2ye
xy
2
+2xy
3
e
xy
2
So, we will find function f(x,y):f(x,y):
\frac{\partial f(x,y)}{\partial x}=y^2e^{xy^2}+4x^3, \ \ \frac{\partial f(x,y) }{\partial y}=2xye^{xy^2}-3y^2
∂x
∂f(x,y)
=y
2
e
xy
2
+4x
3
,
∂y
∂f(x,y)
=2xye
xy
2
−3y
2
f(x,y)=\int (y^2e^{xy^2}+4x^3)dx= e^{xy^2}+x^4+\phi (y)f(x,y)=∫(y
2
e
xy
2
+4x
3
)dx=e
xy
2
+x
4
+ϕ(y)
f(x,y)=\int (2xye^{xy^2}-3y^2)dy =e^{xy^2} -y^3+\psi (x)f(x,y)=∫(2xye
xy
2
−3y
2
)dy=e
xy
2
−y
3
+ψ(x)
f(x,y)=e^{xy^2}+x^4-y^3f(x,y)=e
xy
2
+x
4
−y
3
\frac{\partial f(x,y)}{\partial x} dx+\frac{\partial f(x,y)}{\partial y} dy=0 \ \Rightarrow f(x,y)\equiv C
∂x
∂f(x,y)
dx+
∂y
∂f(x,y)
dy=0 ⇒f(x,y)≡C
Answer: e^{xy^2(x)}+x^4-y^3(x)=Ce
xy
2
(x)
+x
4
−y
3
(x)=C