Question

1. Solve
(x^3+y^3)dx−3xy^2dy=0

2. Solve the variable separable
x^3dx+(y+1)^2dy=0

3. Solve
(1+2e^x/y)dx+2e^x/y(1−x/y)dy=0

4. Solve
y(xy+1)dx+x(1+xy+x^2y^2)dy=0

5. Solve
xdy−ydx−√x^2-y^2 dx=0

Answer 1

Answer on Question #58821 - Math - Differential Equations

Question

1. Solve $(x^{3} + y^{3})dx - 3xy^{2}dy = 0$

Solution

$$(x^{3} + y^{3})dx - 3xy^{2}dy = 0;$$ $$(x^{3} + y^{3})dx = 3xy^{2}dy;$$ $$3xy^{2}dy = (x^{3} + y^{3})dx;$$

Divide both sides by $3xy^{2}dx$:

$$\frac{dy}{dx} = \frac{x^{3} + y^{3}}{3xy^{2}},$$

Divide numerator and denominator by $x^3$.

$$\frac{dy}{dx} = \frac{\frac{x^{3}}{x^{3}} + \frac{y^{3}}{x^{3}}}{\frac{3xy^{2}}{x^{3}}};$$ $$\frac{dy}{dx} = \frac{1 + \frac{y^{3}}{x^{3}}}{\frac{3^{2}}{x^{3}}};$$

Let $u = \frac{y}{x'}$, then $y = ux$, $\frac{dy}{dx} = y' = u'x + u$.

$$u'x + u = \frac{1 + u^{3}}{3u^{2}};$$ $$u'x = \frac{1 + u^{3}}{3u^{2}} - u;$$ $$u'x = \frac{1 + u^{3} - 3u^{3}}{3u^{2}};$$ $$u'x = \frac{1 - 2u^{3}}{3u^{2}};$$ $$\frac{du}{dx}x = \frac{1 - 2u^{3}}{3u^{2}};$$

Separate the variables:

$$\frac{3u^{2}du}{1 - 2u^{3}} = \frac{dx}{x};$$

Integrate both sides:

$$\int \frac{3u^{2}du}{1 - 2u^{3}} = \int \frac{dx}{x};$$ $$-\frac{1}{2} \int \frac{d(1 - 2u^{3})}{1 - 2u^{3}} = \int \frac{dx}{x};$$

$-\frac{1}{2}\ln |1 - 2u^3| = \ln x + \ln C$, where $C$ is an integration constant;

$$\ln \frac {1}{\sqrt {1 - 2 u ^ {3}}} = \ln | C x |.$$ $$\frac {1}{\sqrt {1 - 2 u ^ {3}}} = C x;$$ $$1 - 2 u ^ {3} = \frac {1}{C x ^ {2}};$$ $$2 u ^ {3} = 1 - \frac {1}{C x ^ {2}};$$ $$2 \frac {y ^ {3}}{x ^ {3}} = 1 - \frac {1}{C x ^ {2}};$$ $$y ^ {3} = \frac {x ^ {3}}{2} - \frac {x ^ {3}}{2 C x ^ {2}};$$ $$y ^ {3} = \frac {x ^ {3}}{2} - \frac {x}{2 C}.$$

Then $y = \sqrt[3]{\frac{x^3}{2} - \frac{x}{2C}}$.

Answer: $y = \sqrt[3]{\frac{x^3}{2} - \frac{x}{2C}}$.

Question

2. Solve the variable separable $x^{3}dx + (y + 1)^{2}dy = 0$.

Solution

$$x ^ {3} d x + (y + 1) ^ {2} d y = 0;$$ $$x ^ {3} d x = - (y + 1) ^ {2} d y.$$

Separate the variables:

$$x ^ {3} d x = - (y + 1) ^ {2} d y.$$

Integrate both sides:

$$\int x ^ {3} d x = - \int (y + 1) ^ {2} d y;$$

$\frac{x^4}{4} + C = -\int (y + 1)^2 dy$, where $C$ is an integration constant;

$$- \int (y + 1) ^ {2} d (y + 1) = \frac {x ^ {4}}{4} + C;$$ $$- \frac {(y + 1) ^ {3}}{3} = \frac {x ^ {4}}{4} + C;$$ $$(y + 1) ^ {3} = - \frac {3 x ^ {4}}{4} - 3 C;$$ $$y + 1 = \sqrt[3]{-\frac{3x^4}{4} - 3C};$$ $$y = -1 + \sqrt[3]{-\frac{3x^4}{4} - 3C}.$$

Answer: $y = -1 + \sqrt[3]{-\frac{3x^4}{4} - 3C}$.

Question

3. Solve $\left(1 + 2e^{\frac{x}{y}}\right)dx + 2e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy = 0$.

Solution

$$\left(1 + 2e^{\frac{x}{y}}\right)dx + 2e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy = 0.$$

Let $P(x,y) = 1 + 2e^{\frac{x}{y}}$, $Q(x,y) = 2e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right) = 2e^{\frac{x}{y}} - 2e^{\frac{x}{y}}x$.

Find $\frac{\partial P(x,y)}{\partial y} = -\frac{2xe^{\frac{x}{y}}}{y^2}$ and $\frac{\partial Q(x,y)}{\partial x} = -\frac{2xe^{\frac{x}{y}}}{y^2}$. Because $\frac{\partial P(x,y)}{\partial y} = \frac{\partial Q(x,y)}{\partial x}$, we deal with an exact equation.

Define $f(x,y)$ such that $\frac{\partial f(x,y)}{\partial x} = P(x,y)$ and $\frac{\partial f(x,y)}{\partial y} = Q(x,y)$.

The solution is $f(x,y) = C$, where $C$ is an arbitrary real constant;

$$f(x,y) = \int P(x,y)dx + g(y) = \int \left(1 + 2e^{\frac{x}{y}}\right)dx + g(y) = x + 2ye^{\frac{x}{y}} + g(y).$$ $$f(x,y) = \int Q(x,y)dy + r(x) = \int 2e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)dy + r(x) = 2ye^{\frac{x}{y}} + r(x).$$

Equate both expressions for $f(x,y)$:

$$x + 2ye^{\frac{x}{y}} + g(y) = 2ye^{\frac{x}{y}} + r(x),$$

hence $g(y) = 0$ and $r(x) = x$, therefore $f(x,y) = x + 2ye^{\frac{x}{y}} = C$.

Answer: $x + 2ye^{\frac{x}{y}} = C$.

Question

4. Solve $y(xy + 1)dx + x(1 + xy + x^2 y^2)dy = 0$

Solution

Divide both sides by $dx$:

$$y(xy + 1) + x(1 + xy + x^2 y^2) \frac{dy}{dx} = 0;$$

Let $u = xy$. Then $y = \frac{u}{x'}$. $\frac{dy}{dx} = y' = \frac{u'}{x} - \frac{u}{x^2}$;

$$\frac{u}{x}(u + 1) + x(1 + u + u^2) \left(\frac{u'}{x} - \frac{u}{x^2}\right) = 0;$$

Multiply both sides by $x$:

$$u^2 + u + (1 + u + u^2)(u'x - u) = 0;$$ $$u^2 + u + u'x(1 + u + u^2) - u - u^2 - u^3 = 0;$$ $$u'x(1 + u + u^2) - u^3 = 0;$$ $$u'x(1 + u + u^2) = u^3;$$ $$\frac{du}{dx} x(1 + u + u^2) = u^3.$$

Separate the variables:

$$\frac{(1 + u + u^2)du}{u^3} = \frac{dx}{x}.$$

Integrate both sides:

$$\int \frac{(1 + u + u^2)du}{u^3} = \int \frac{dx}{x};$$ $$\int u^{-3} du + \int u^{-2} du + \int \frac{du}{u} = \ln x + C;$$

$C$ is an integration constant;

$$\frac{1}{-2u^2} - \frac{1}{u} + \ln u = \ln x + C.$$

Go back to $u = xy$, then find $\frac{1}{-2x^2y^2} - \frac{1}{xy} + \ln(xy) = \ln x + C$;

$$\frac{1}{-2x^2y^2} - \frac{1}{xy} + \ln x + \ln y = \ln x + C;$$ $$\frac{1}{-2x^2y^2} - \frac{1}{xy} + \ln y = C.$$

Answer: $-\frac{1}{2x^2y^2} - \frac{1}{xy} + \ln y = C.$

Question

5. Solve $xdy - ydx - \sqrt{x^2 - y^2} dx = 0$

Solution

$$xdy - ydx - \sqrt{x^2 - y^2} dx = 0;$$ $$xdy = (y + \sqrt{x^2 - y^2}) dx.$$

Divide both sides by $xdx$:

$$\frac{dy}{dx} = \frac{(y + \sqrt{x^2 - y^2})}{x},$$ $$\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 - \frac{y^2}{x^2}};$$

Let $u = \frac{y}{x'}$ then $y = ux$, $\frac{dy}{dx} = y' = u'x + u$.

$$u'x + u = u + \sqrt{1 - u^2};$$ $$u'x = \sqrt{1 - u^2};$$ $$\frac{du}{dx} x = \sqrt{1 - u^2}.$$

Separate the variables:

$$\frac{du}{\sqrt{1 - u^2}} = \frac{dx}{x}.$$

Integrate both sides:

$$\int \frac{du}{\sqrt{1 - u^2}} = \int \frac{dx}{x};$$ $$\sin^{-1} u = \ln x + \ln C;$$

$C$ is an integration constant.

$$u = \sin(\ln Cx).$$

Go back to $u = \frac{y}{x}$, then $\frac{y}{x} = \sin(\ln Cx)$,

$$y = x \sin(\ln Cx).$$

Answer: $y = x \sin(\ln Cx)$.

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