Question

Solve the differential equation
1.sin^-1(dy/dx)=(x+y)
2.(1+y^2)dx=(tan^-1y-x)dy
3.(D-1)^2(D^2+1)^2y=sin^2(x/2)+e^x+x
4.2x^2y(d^2y/dx^2)+4y^2=x^2(dy/dx)^2+2xy(dy/dx)

Answer 1

Answer on Question #76036 - Math - Differential Equations

Solve the differential equation

Question

1.

$$\sin^{-1} \left(\frac{dy}{dx}\right) = x + y$$

Solution

$$\frac{dy}{dx} = \sin(x + y)$$ $$x + y = u$$ $$1 + \frac{dy}{dx} = \frac{du}{dx}$$ $$\frac{dy}{dx} = \frac{du}{dx} - 1$$ $$\frac{du}{dx} - 1 = \sin u$$ $$\frac{du}{dx} = \sin u + 1$$ $$\int \frac{du}{\sin u + 1} = \int dx = x + c$$ $$\int \frac{du}{\sin u + 1} = \int \frac{1 - \sin u}{1 - \sin^2 u} \, du = \int \frac{du}{\cos^2 u} - \int \frac{\sin u}{\cos^2 u} \, du$$ $$\int \frac{\sin u}{\cos^2 u} \, du = - \int \frac{d(\cos u)}{\cos^2 u} = \frac{1}{\cos u}$$ $$\int \frac{du}{\cos^2 u} = \int (1 + \tan^2 u) \, du \Rightarrow t = \tan u, \, dt = (1 + t^2) \, du, \, du = \frac{dt}{1 + t^2}$$ $$\int \frac{du}{\cos^2 u} = \int \frac{1 + t^2}{1 + t^2} \, dt = t = \tan u$$ $$\tan u - \frac{1}{\cos u} = x + c$$

Answer:

$$\tan (x + y) - \frac {1}{\cos (x + y)} = x + c$$

Question

2.

$$(1 + y ^ {2}) d x = (\tan^ {- 1} y - x) d y$$

Solution

$$\frac {d x}{d y} = \frac {\tan^ {- 1} y - x}{1 + y ^ {2}}$$ $$\frac {d x}{d y} + \frac {x}{1 + y ^ {2}} = \frac {\tan^ {- 1} y}{1 + y ^ {2}}$$

Integrating factor:

$$e ^ {\int \frac {1}{1 + y ^ {2}} d y} = e ^ {\tan^ {- 1} y}$$ $$x e ^ {\tan^ {- 1} y} = \int \frac {\tan^ {- 1} y}{1 + y ^ {2}} e ^ {\tan^ {- 1} y} d y$$ $$t = \tan^ {- 1} y$$ $$d t = \frac {1}{1 + y ^ {2}} d y$$ $$x e ^ {\tan^ {- 1} y} = \int t e ^ {t} d t = t e ^ {t} - e ^ {t} + c = e ^ {t} (t - 1) + c$$ $$x e ^ {\tan^ {- 1} y} = e ^ {\tan^ {- 1} y} (\tan^ {- 1} y - 1) + c$$

Answer:

$$x = \tan^ {- 1} y - 1 + c e ^ {- \tan^ {- 1} y}$$

Question

3.

$$(D - 1) ^ {2} (D ^ {2} + 1) ^ {2} y = \left(\sin \frac {x}{2}\right) ^ {2} + e ^ {x} + x$$

Solution

$$\left(\sin \frac {x}{2}\right) ^ {2} = \frac {1 - \cos x}{2}$$ $$(D - 1) ^ {2} (D ^ {2} + 1) ^ {2} y = \frac {1 - \cos x}{2} + e ^ {x} + x$$ $$(y ^ {\prime} - y) ^ {2} (y ^ {\prime \prime} + y) ^ {2} = 0$$

Auxiliary equation:

$$(\lambda - 1) ^ {2} (\lambda^ {2} + 1) ^ {2} = 0$$ $$\lambda_ {1, 2} = 1, \lambda_ {3, 4} = i, \lambda_ {5, 6} = - i$$

General solution:

$$Y = c _ {1} e ^ {x} + c _ {2} x e ^ {x} + c _ {3} \cos x + c _ {4} \sin x + c _ {5} x \cos x + c _ {6} x \sin x$$

Particular integral:

$$\tilde {y} = \tilde {y} _ {1} + \tilde {y} _ {2} + \tilde {y} _ {3}$$

Particular integral $\tilde{y}_1$ for:

$$(y ^ {\prime} - y) ^ {2} (y ^ {\prime \prime} + y) ^ {2} = \frac {1}{2} + x$$ $$\tilde {y} _ {1} = A + B x$$ $$\tilde {y} _ {1} ^ {\prime} = B$$ $$\tilde {y} _ {1} ^ {\prime \prime} = 0$$ $$(B - A - B x) ^ {2} (A + B x) ^ {2} = \frac {1}{2} + x$$

We cannot find coefficients $A, B$ using analytical methods.

Particular integral $\tilde{y}_2$ for:

$$(y' - y)^2(y'' + y)^2 = e^x$$ $$\tilde{y}_2 = A x^2 e^x$$ $$\tilde{y}_2' = 2A x e^x + A x^2 e^x$$ $$\tilde{y}_2'' = 2A x e^x + 2A e^x + 2A x e^x + A x^2 e^x = 2A e^x + 4A x e^x + A x^2 e^x$$ $$(2A x e^x + A x^2 e^x - A x^2 e^x)^2 (2A e^x + 4A x e^x + A x^2 e^x + A x^2 e^x)^2 = e^x$$ $$4A^2 x^2 e^{2x} (2A e^x + 4A x e^x + 2A x^2 e^x)^2 = e^x$$

We cannot find coefficient $A$ using analytical methods.

Particular integral $\tilde{y}_3$ for:

$$(y' - y)^2(y'' + y)^2 = -\frac{1}{2} \cos x$$ $$\tilde{y}_3 = A x^2 \cos x + B x^2 \sin x$$ $$\tilde{y}_3' = 2x (A \cos x + B \sin x) + x^2 (B \cos x - A \sin x)$$ $$\tilde{y}_3'' = 2(A \cos x + B \sin x) + 4x (B \cos x - A \sin x) - x^2 (A \cos x + B \sin x)$$ $$\begin{array}{l} \left((A \cos x + B \sin x)(2x - x^2) + x^2 (B \cos x - A \sin x)\right)^2 \left(2(A \cos x + B \sin x) + 4x (B \cos x - A \sin x)\right)^2 = \\ = -\frac{1}{2} \cos x \end{array}$$

We cannot find coefficients $A, B$ using analytical methods.

**Answer**: It is impossible to solve the given equation using analytical methods, and for numerical methods we have not initial value.

We can use, for example, Runge-Kutta method (if we have initial value) which uses iterations to find value $y(x)$ on the defined interval.

Question

4.

$$2 x ^ {2} y \frac {d ^ {2} y}{d x ^ {2}} + 4 y ^ {2} = x ^ {2} \left(\frac {d y}{d x}\right) ^ {2} + 2 x y \frac {d y}{d x}$$

**Solution**

$$u = \frac {y _ {x}}{y}$$ $$u _ {x} + \frac {u ^ {2}}{2} - \frac {u}{x} + \frac {2}{x ^ {2}} = 0$$ $$v (x) = e ^ {\frac {1}{2} \int u (x) d x}$$ $$x ^ {2} v _ {x x} - x v _ {x} + v = 0$$ $$v (x) = | x | \left(c _ {1} + c _ {2} \ln | x |\right)$$ $$\ln v = \frac {1}{2} \int u (x) d x$$ $$\frac {v _ {x}}{v} = \frac {u}{2}$$ $$v _ {x} = c _ {1} + c _ {2} + c _ {2} \ln x$$ $$\frac {c _ {1} + c _ {2} + c _ {2} \ln x}{x (c _ {1} + c _ {2} \ln x)} = \frac {y _ {x}}{2 y}$$ $$\frac {c _ {1} + c _ {2} + c _ {2} \ln x}{x (c _ {1} + c _ {2} \ln x)} d x = \frac {d y}{2 y}$$ $$\int \frac {c _ {1} + c _ {2} + c _ {2} \ln x}{x (c _ {1} + c _ {2} \ln x)} d x = \int \frac {d y}{2 y}$$

**Answer:**

$$y = c _ {2} x ^ {2} (c _ {1} + \ln x) ^ {2}$$

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