Mister Exam
Other calculators
- 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 x^4(dy/dx)+x^3y=-sec(xy)
- Equation x^2y''+4xy'+2y=e^x
- Equation y'+2sin(2pix)=0
- Equation xy'-3y=6
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Identical expressions
- x^ four (dy/dx)+x^3y=-sec(xy)
- x to the power of 4(dy divide by dx) plus x cubed y equally minus sec(xy)
- x to the power of four (dy divide by dx) plus x cubed y equally minus sec(xy)
- x4(dy/dx)+x3y=-sec(xy)
- x4dy/dx+x3y=-secxy
- x⁴(dy/dx)+x³y=-sec(xy)
- x to the power of 4(dy/dx)+x to the power of 3y=-sec(xy)
- x^4dy/dx+x^3y=-secxy
- x^4(dy divide by dx)+x^3y=-sec(xy)
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Similar expressions
- x^4(dy/dx)+x^3y=+sec(xy)
- x^4(dy/dx)-x^3y=-sec(xy)
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Expressions with functions
- dy
- dy=(x-3y)dx
- dy/dx=(x+3y)/(3x+y)
- dy/dx=4y-8
- dy/dx=ycosx/1+2y^2
- dy/dx=xy^(1/2)
- dx
- dx+xydy=(y^2)dx+ydy
- dx/dt=ax-bx^2
- dx/dt=r-kx
- dx/dt+x/t^2=1/t^2
- dx/dt=3xt^2
- sec
- sec^2xtanydx+sec^2ytan(x)dy=0
- sec^2(x)tanydx+sec^2(y)tanxdy=0
- sec^2*tgy*dx+sec^2*tgx*dy=0
- sec^2xtgydx+sec^2ytgxdy=0
- sec^2*tgydx+sec^2ytgxdy=0
- xy
- xy'-3y=6
- xydx-(x+2)dy=0
- xydx+(x^2+y^2)dy=0
- xydx-(x+2y)^2dy=0
- xy^3dx+e^(x^2)dy=0
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What you mean?
- x^3*y + dy*x^4/dx = -sec(x*y)
- x^(3*y) + dy*x^4/dx = -sec(x*y)
- x^(3*y) + dy*x^4/dx = -sec(x*y)
You entered:
x^4(dy/dx)+x^3y=-sec(xy)
What you mean?
Differential equation x^4(dy/dx)+x^3y=-sec(xy)
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The solution
You have entered
[src]
3 4 d
x *y(x) + x *--(y(x)) = -sec(x*y(x))
dx
$$x^{4} \frac{d}{d x} y{\left(x \right)} + x^{3} y{\left(x \right)} = - \sec{\left(x y{\left(x \right)} \right)}$$
x^4*y' + x^3*y = -sec(x*y)
Detail solution
Given the equation:
$$x^{4} \frac{d}{d x} y{\left(x \right)} + x^{3} y{\left(x \right)} + \sec{\left(x y{\left(x \right)} \right)} = 0$$
Do replacement
$$u{\left(x \right)} = x y{\left(x \right)}$$
and because
$$y{\left(x \right)} = \frac{u{\left(x \right)}}{x}$$
then
$$\frac{d}{d x} y{\left(x \right)} = \frac{\frac{d}{d x} u{\left(x \right)}}{x} - \frac{u{\left(x \right)}}{x^{2}}$$
substitute
$$x^{4} \frac{d}{d x} \frac{u{\left(x \right)}}{x} + x^{2} u{\left(x \right)} + \sec{\left(u{\left(x \right)} \right)} = 0$$
or
$$x^{3} \frac{d}{d x} u{\left(x \right)} + \sec{\left(u{\left(x \right)} \right)} = 0$$
This differential equation has the form:
where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(u \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \frac{1}{x^{3}}$$
$$\operatorname{g_{2}}{\left(u \right)} = \sec{\left(u{\left(x \right)} \right)}$$
We give the equation to the form:
Divide both parts of the equation by g2(u)
$$\sec{\left(u{\left(x \right)} \right)}$$
we get
$$\cos{\left(u{\left(x \right)} \right)} \frac{d}{d x} u{\left(x \right)} = - \frac{1}{x^{3}}$$
We separated the variables x and u.
Now, multiply the both equation sides by dx,
then the equation will be the
$$dx \cos{\left(u{\left(x \right)} \right)} \frac{d}{d x} u{\left(x \right)} = - \frac{dx}{x^{3}}$$
or
$$du \cos{\left(u{\left(x \right)} \right)} = - \frac{dx}{x^{3}}$$
Take the integrals from the both equation sides:
- the integral of the left side by u,
- the integral of the right side by x.
$$\int \cos{\left(u \right)}\, du = \int \left(- \frac{1}{x^{3}}\right)\, dx$$
Detailed solution of the integral with u
Detailed solution of the integral with x
Take this integrals
$$\sin{\left(u \right)} = Const + \frac{1}{2 x^{2}}$$
Detailed solution of the equation
We get the simple equation with the unknown variable u.
(Const - it is a constant)
Solution is:
$$\operatorname{u_{1}} = u{\left(x \right)} = \pi - \operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}$$
$$\operatorname{u_{2}} = u{\left(x \right)} = \operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}$$
do backward replacement
$$y{\left(x \right)} = \frac{u{\left(x \right)}}{x}$$
$$y1 = y(x) = \frac{\pi - \operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}}{x}$$
$$y2 = y(x) = \frac{\operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}}{x}$$
$$x^{4} \frac{d}{d x} y{\left(x \right)} + x^{3} y{\left(x \right)} + \sec{\left(x y{\left(x \right)} \right)} = 0$$
Do replacement
$$u{\left(x \right)} = x y{\left(x \right)}$$
and because
$$y{\left(x \right)} = \frac{u{\left(x \right)}}{x}$$
then
$$\frac{d}{d x} y{\left(x \right)} = \frac{\frac{d}{d x} u{\left(x \right)}}{x} - \frac{u{\left(x \right)}}{x^{2}}$$
substitute
$$x^{4} \frac{d}{d x} \frac{u{\left(x \right)}}{x} + x^{2} u{\left(x \right)} + \sec{\left(u{\left(x \right)} \right)} = 0$$
or
$$x^{3} \frac{d}{d x} u{\left(x \right)} + \sec{\left(u{\left(x \right)} \right)} = 0$$
This differential equation has the form:
f1(x)*g1(u)*u' = f2(x)*g2(u),
where
$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(u \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \frac{1}{x^{3}}$$
$$\operatorname{g_{2}}{\left(u \right)} = \sec{\left(u{\left(x \right)} \right)}$$
We give the equation to the form:
g1(u)/g2(u)*u'= f2(x)/f1(x).
Divide both parts of the equation by g2(u)
$$\sec{\left(u{\left(x \right)} \right)}$$
we get
$$\cos{\left(u{\left(x \right)} \right)} \frac{d}{d x} u{\left(x \right)} = - \frac{1}{x^{3}}$$
We separated the variables x and u.
Now, multiply the both equation sides by dx,
then the equation will be the
$$dx \cos{\left(u{\left(x \right)} \right)} \frac{d}{d x} u{\left(x \right)} = - \frac{dx}{x^{3}}$$
or
$$du \cos{\left(u{\left(x \right)} \right)} = - \frac{dx}{x^{3}}$$
Take the integrals from the both equation sides:
- the integral of the left side by u,
- the integral of the right side by x.
$$\int \cos{\left(u \right)}\, du = \int \left(- \frac{1}{x^{3}}\right)\, dx$$
Detailed solution of the integral with u
Detailed solution of the integral with x
Take this integrals
$$\sin{\left(u \right)} = Const + \frac{1}{2 x^{2}}$$
Detailed solution of the equation
We get the simple equation with the unknown variable u.
(Const - it is a constant)
Solution is:
$$\operatorname{u_{1}} = u{\left(x \right)} = \pi - \operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}$$
$$\operatorname{u_{2}} = u{\left(x \right)} = \operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}$$
do backward replacement
$$y{\left(x \right)} = \frac{u{\left(x \right)}}{x}$$
$$y1 = y(x) = \frac{\pi - \operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}}{x}$$
$$y2 = y(x) = \frac{\operatorname{asin}{\left(C_{1} + \frac{1}{2 x^{2}} \right)}}{x}$$
Graph of the Cauchy problem
Numerical answer
[src]
(x, y):
(-10.0, 0.75)
(-7.777777777777778, 0.9630896691322584)
(-5.555555555555555, 1.344417613469695)
(-3.333333333333333, 2.219516965400296)
(-1.1111111111111107, 6.1661246695118255)
(1.1111111111111107, 9.289501605317518)
(3.333333333333334, 3.1933833808213433e-248)
(5.555555555555557, 4.192407976641303e+175)
(7.777777777777779, 8.388243567338859e+296)
(10.0, 3.861029683e-315)
(10.0, 3.861029683e-315)
The graph