Solve Differential equation
(1+2xycosx²-2xy)dx+(sinx²-x²)dy=0
Answers
Step-by-step explanation:
1+2xycosx
2
−2xy)dx+(sinx
2
−x
2
)dy=0
\left(1+2xy\cos \left(x^2\right)-2xy\right)dx+\left(\sin \left(x^2\right)-x^2\right)dy=0(1+2xycos(x
2
)−2xy)dx+(sin(x
2
)−x
2
)dy=0
First order linear Ordinary Differential Equation
\mathrm{A\:first\:order\:linear\:ODE\:has\:the\:form\:of\:}y'\left(x\right)+p\left(x\right)y=q\left(x\right)AfirstorderlinearODEhastheformofy
′
(x)+p(x)y=q(x)
Let y be the dependent variable, Divide by dx:
1+2xy\cos \left(x^2\right)-2xy+\left(\sin \left(x^2\right)-x^2\right)\frac{dy}{dx}=01+2xycos(x
2
)−2xy+(sin(x
2
)−x
2
)
dx
dy
=0
\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'Substitute
dx
dy
withy
′
1+2xy\cos \left(x^2\right)-2xy+\left(\sin \left(x^2\right)-x^2\right)y'\:=01+2xycos(x
2
)−2xy+(sin(x
2
)−x
2
)y
′
=0
We rewrite in the form of a first order linear ODE
y'\:+\frac{2x\left(\cos \left(x^2\right)-1\right)}{-x^2+\sin \left(x^2\right)}y=-\frac{1}{\sin \left(x^2\right)-x^2}y
′
+
−x
2
+sin(x
2
)
2x(cos(x
2
)−1)
y=−
sin(x
2
)−x
2
1
We find integration factor: \mu = -x^2 + \sin \left(x^2\right)μ=−x
2
+sin(x
2
)
We put the equation in the from (\mu(x)⋅y)' = \mu(x) ⋅ q(x); ((-x^2 + sin(x^2))y)' = -1(μ(x)⋅y)
′
=μ(x)⋅q(x);((−x
2
+sin(x
2
))y)
′
=−1
We solve:
((-x^2 + sin(x^2))y)' = -1;\, y=-\frac{x}{-x^2+\sin \left(x^2\right)}+\frac{c_1}{-x^2+\sin \left(x^2\right)}((−x
2
+sin(x
2
))y)
′
=−1;y=−
−x
2
+sin(x
2
)
x
+
−x
2
+sin(x
2
)
c
1
y=-\frac{x}{-x^2+\sin \left(x^2\right)}+\frac{c_1}{-x^2+\sin \left(x^2\right)}y=−
−x
2
+sin(x
2
)
x
+
−x
2
+sin(x
2
)
c
1