Answer to Question #260747 in Differential Equations for christian

Question

dy/dx + y/x =y^-2

dy/dx +xy=x sqrt y

Answer 1

a. Rewrite in the form of a first order Bernoulli ODE

$$\dfrac{dy}{dx}+\dfrac{1}{x}y=y^{-2}$$

Substitution $z=y^{1-(-2)}=y^3$

$$\dfrac{dz}{dx}=3y^2\dfrac{dy}{dx}$$

$$y^2\dfrac{dy}{dx}+\dfrac{1}{x}y^3=1$$

$$\dfrac{1}{3}\dfrac{dz}{dx}+\dfrac{1}{x}z=1$$

$$\dfrac{dz}{dx}+\dfrac{3}{x}z=3$$

Integrating factor

$$\mu(x)=e^{\int(3/x)dx}=x^3$$

$$x^3\dfrac{dz}{dx}+3x^2z=3x^3$$

$$d(x^3z)=3x^3dx$$

Integrate

$$\int d(x^3z)=\int 3x^3dx$$

$$x^3z=\dfrac{3}{4}x^4+C$$

$$z=\dfrac{3}{4}x+\dfrac{C}{x^3}$$

$$y=\sqrt[3]{\dfrac{3}{4}x+\dfrac{C}{x^3}}$$

b. The equation is in the form of a first order Bernoulli ODE

$$\dfrac{dy}{dx}+xy=xy^{1/2}$$

Substitution $z=y^{1-(1/2)}=y^{1/2}$

$$\dfrac{dz}{dx}=\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}$$

$$\dfrac{1}{2\sqrt{y}}\dfrac{dy}{dx}+\dfrac{1}{2}x\sqrt{y}=\dfrac{1}{2}x$$

$$\dfrac{dz}{dx}+\dfrac{1}{2}xz=\dfrac{1}{2}x$$

Integrating factor

$$\mu(x)=e^{\int(x/2)dx}=e^{x^2/4}$$

$$e^{x^2/4}\dfrac{dz}{dx}+\dfrac{1}{2}xe^{x^2/4}z=\dfrac{1}{2}xe^{x^2/4}$$

$$d(e^{x^2/4}z)=\dfrac{1}{2}xe^{x^2/4}dx$$

Integrate

$$\int d(e^{x^2/4}z)=\int \dfrac{1}{2}xe^{x^2/4}dx$$

$$e^{x^2/4}z=e^{x^2/4}+C$$

$$z=1+Ce^{-x^2/4}$$

$$y=(1+Ce^{-x^2/4})^2$$