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Differential equation x^4(dy/dx)+x^3y=-sec(xy)

For Cauchy problem:

y() =
y'() =
y''() =
y'''() =
y''''() =

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The solution

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 3         4 d                      
x *y(x) + x *--(y(x)) = -sec(x*y(x))
$$x^{4} \frac{d}{d x} y{\left(x \right)} + x^{3} y{\left(x \right)} = - \sec{\left(x y{\left(x \right)} \right)}$$
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
$$y{\left(x \right)} = \frac{u{\left(x \right)}}{x}$$
$$\frac{d}{d x} y{\left(x \right)} = \frac{\frac{d}{d x} u{\left(x \right)}}{x} - \frac{u{\left(x \right)}}{x^{2}}$$
$$x^{4} \frac{d}{d x} \frac{u{\left(x \right)}}{x} + x^{2} u{\left(x \right)} + \sec{\left(u{\left(x \right)} \right)} = 0$$
$$x^{3} \frac{d}{d x} u{\left(x \right)} + \sec{\left(u{\left(x \right)} \right)} = 0$$
This differential equation has the form:
f1(x)*g1(u)*u' = f2(x)*g2(u),

$$\operatorname{f_{1}}{\left(x \right)} = 1$$
$$\operatorname{g_{1}}{\left(u \right)} = 1$$
$$\operatorname{f_{2}}{\left(x \right)} = - \frac{1}{x^{3}}$$
$$\operatorname{g_{2}}{\left(u \right)} = \sec{\left(u{\left(x \right)} \right)}$$
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
$$dx \cos{\left(u{\left(x \right)} \right)} \frac{d}{d x} u{\left(x \right)} = - \frac{dx}{x^{3}}$$
$$du \cos{\left(u{\left(x \right)} \right)} = - \frac{dx}{x^{3}}$$

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
Differential equation x^4(dy/dx)+x^3y=-sec(xy)
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