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- Linear homogeneous differential equations of 2nd order Step-By-Step
- Linear inhomogeneous differential equations of the 1st order Step-By-Step
- Differential equations with separable variables Step-by-Step
- A simplest differential equations of 1-order Step-by-Step
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How to use it?
- Differential equation:
- Equation xy''-(2x+1)y'+(x+1)y=0
- Equation dy/dx=(x^2+3y^2)/(2xy)
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Equation y'+4y=1.4
- Equation xe^ydy+x^2+1/ydx=0
- Derivative of:
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dy/dx
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Identical expressions
- dy/dx=(x^ two +3y^ two)/(2xy)
- dy divide by dx equally (x squared plus 3y squared ) divide by (2xy)
- dy divide by dx equally (x to the power of two plus 3y to the power of two) divide by (2xy)
- dy/dx=(x2+3y2)/(2xy)
- dy/dx=x2+3y2/2xy
- dy/dx=(x²+3y²)/(2xy)
- dy/dx=(x to the power of 2+3y to the power of 2)/(2xy)
- dy/dx=x^2+3y^2/2xy
- dy divide by dx=(x^2+3y^2) divide by (2xy)
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Similar expressions
- dy/dx=(x^2-3y^2)/(2xy)
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Expressions with functions
- dy
- dy/dx-y=e^xy^2
- dy/dx=4y-8
- dy/dx=5y
- dy/dx=ycosx/1+2y^2
- dy/dx+y^2sinx=0
- dx
- dx/dt=x
- dx/dt=3xt^2
- dx/dt=-k(a-x)
- dx/dt+3x/20+t=4
- dx/dt=kx(n+1-x)
Differential equation dy/dx=(x^2+3y^2)/(2xy)
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The solution
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[src]
2 2 d x + 3*y (x) --(y(x)) = ------------ dx 2*x*y(x)
$$\frac{d}{d x} y{\left(x \right)} = \frac{x^{2} + 3 y^{2}{\left(x \right)}}{2 x y{\left(x \right)}}$$
y' = (x^2 + 3*y^2)/(2*x*y)
Detail solution
Given the equation:
$$\frac{d}{d x} y{\left(x \right)} - \frac{x^{2} + 3 y^{2}{\left(x \right)}}{2 x y{\left(x \right)}} = 0$$
Do replacement
$$u{\left(x \right)} = \frac{y{\left(x \right)}}{x}$$
and because
$$y{\left(x \right)} = x u{\left(x \right)}$$
then
$$\frac{d}{d x} y{\left(x \right)} = x \frac{d}{d x} u{\left(x \right)} + u{\left(x \right)}$$
substitute
$$- \frac{3 u{\left(x \right)}}{2} + \frac{d}{d x} x u{\left(x \right)} - \frac{1}{2 u{\left(x \right)}} = 0$$
or
$$x \frac{d}{d x} u{\left(x \right)} - \frac{u{\left(x \right)}}{2} - \frac{1}{2 u{\left(x \right)}} = 0$$
This differential equation has the form:
$$f_1(x)*g_1(u)*u' = f_2(x)*g_2(u)$$
where
$$f_{1}{\left(x \right)} = 1$$
$$g_{1}{\left(u \right)} = 1$$
$$f_{2}{\left(x \right)} = - \frac{1}{x}$$
$$g_{2}{\left(u \right)} = - \frac{u^{2}{\left(x \right)} + 1}{2 u{\left(x \right)}}$$
We give the equation to the form:
$$\frac{g_1(u)}{g_2(u)}*u'= \frac{f_2(x)}{f_1(x)}$$
Divide both parts of the equation by g_2(u)
$$- \frac{u^{2}{\left(x \right)} + 1}{2 u{\left(x \right)}}$$
we get
$$- \frac{2 u{\left(x \right)} \frac{d}{d x} u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{1}{x}$$
We separated the variables x and u.
Now, multiply the both equation sides by dx,
then the equation will be the
$$- \frac{2 dx u{\left(x \right)} \frac{d}{d x} u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{dx}{x}$$
or
$$- \frac{2 du u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{dx}{x}$$
Take the integrals from the both equation sides:
- the integral of the left side by u,
- the integral of the right side by x.
$$\int \left(- \frac{2 u}{u^{2} + 1}\right)\, du = \int \left(- \frac{1}{x}\right)\, dx$$
Detailed solution of the integral with u
Detailed solution of the integral with x
Take this integrals
$$- \log{\left(u^{2} + 1 \right)} = Const - \log{\left(x \right)}$$
Detailed solution of the equation
We get the simple equation with the unknown variable u.
(Const - it is a constant)
Solution is:
$$u_{1} = u{\left(x \right)} = - \sqrt{C_{1} x - 1}$$
$$u_{2} = u{\left(x \right)} = \sqrt{C_{1} x - 1}$$
do backward replacement
$$y{\left(x \right)} = x u{\left(x \right)}$$
$$y_{1} = y{\left(x \right)} = - x \sqrt{C_{1} x - 1}$$
$$y_{2} = y{\left(x \right)} = x \sqrt{C_{1} x - 1}$$
$$\frac{d}{d x} y{\left(x \right)} - \frac{x^{2} + 3 y^{2}{\left(x \right)}}{2 x y{\left(x \right)}} = 0$$
Do replacement
$$u{\left(x \right)} = \frac{y{\left(x \right)}}{x}$$
and because
$$y{\left(x \right)} = x u{\left(x \right)}$$
then
$$\frac{d}{d x} y{\left(x \right)} = x \frac{d}{d x} u{\left(x \right)} + u{\left(x \right)}$$
substitute
$$- \frac{3 u{\left(x \right)}}{2} + \frac{d}{d x} x u{\left(x \right)} - \frac{1}{2 u{\left(x \right)}} = 0$$
or
$$x \frac{d}{d x} u{\left(x \right)} - \frac{u{\left(x \right)}}{2} - \frac{1}{2 u{\left(x \right)}} = 0$$
This differential equation has the form:
$$f_1(x)*g_1(u)*u' = f_2(x)*g_2(u)$$
where
$$f_{1}{\left(x \right)} = 1$$
$$g_{1}{\left(u \right)} = 1$$
$$f_{2}{\left(x \right)} = - \frac{1}{x}$$
$$g_{2}{\left(u \right)} = - \frac{u^{2}{\left(x \right)} + 1}{2 u{\left(x \right)}}$$
We give the equation to the form:
$$\frac{g_1(u)}{g_2(u)}*u'= \frac{f_2(x)}{f_1(x)}$$
Divide both parts of the equation by g_2(u)
$$- \frac{u^{2}{\left(x \right)} + 1}{2 u{\left(x \right)}}$$
we get
$$- \frac{2 u{\left(x \right)} \frac{d}{d x} u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{1}{x}$$
We separated the variables x and u.
Now, multiply the both equation sides by dx,
then the equation will be the
$$- \frac{2 dx u{\left(x \right)} \frac{d}{d x} u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{dx}{x}$$
or
$$- \frac{2 du u{\left(x \right)}}{u^{2}{\left(x \right)} + 1} = - \frac{dx}{x}$$
Take the integrals from the both equation sides:
- the integral of the left side by u,
- the integral of the right side by x.
$$\int \left(- \frac{2 u}{u^{2} + 1}\right)\, du = \int \left(- \frac{1}{x}\right)\, dx$$
Detailed solution of the integral with u
Detailed solution of the integral with x
Take this integrals
$$- \log{\left(u^{2} + 1 \right)} = Const - \log{\left(x \right)}$$
Detailed solution of the equation
We get the simple equation with the unknown variable u.
(Const - it is a constant)
Solution is:
$$u_{1} = u{\left(x \right)} = - \sqrt{C_{1} x - 1}$$
$$u_{2} = u{\left(x \right)} = \sqrt{C_{1} x - 1}$$
do backward replacement
$$y{\left(x \right)} = x u{\left(x \right)}$$
$$y_{1} = y{\left(x \right)} = - x \sqrt{C_{1} x - 1}$$
$$y_{2} = y{\left(x \right)} = x \sqrt{C_{1} x - 1}$$
The answer
[src]
___________ y(x) = -x*\/ -1 + C1*x
$$y{\left(x \right)} = - x \sqrt{C_{1} x - 1}$$
___________ y(x) = x*\/ -1 + C1*x
$$y{\left(x \right)} = x \sqrt{C_{1} x - 1}$$
The answer (#2)
[src]
$$y\left(x\right)={\it ilt}\left({{\mathcal{L}\left({{3\,y\left(x
\right)}\over{2\,x}}+{{x}\over{2\,y\left(x\right)}} , x , g_{19164}
\right)+y\left(0\right)}\over{g_{19164}}} , g_{19164} , x\right)$$
y = 'ilt(('laplace((3*y)/(2*x)+x/(2*y),x,g19164)+y(0))/g19164,g19164,x)
Graph of the Cauchy problem
The classification
1st homogeneous coeff best
1st homogeneous coeff subs dep div indep
1st homogeneous coeff subs dep div indep Integral
1st homogeneous coeff subs indep div dep
1st homogeneous coeff subs indep div dep Integral
Bernoulli
Bernoulli Integral
factorable
lie group
Numerical answer
[src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 4.025199483059684e-10)
(-5.555555555555555, 2.17e-322)
(-3.333333333333333, nan)
(-1.1111111111111107, 2.78363573e-315)
(1.1111111111111107, 8.427456047434801e+197)
(3.333333333333334, 3.1933833808213433e-248)
(5.555555555555557, 1.0400058166471786e-42)
(7.777777777777779, 8.388243567338858e+296)
(10.0, 9.527767843605595e-305)
(10.0, 9.527767843605595e-305)