Question

The differential equation $\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}\mathbf{=}\mathit{P}\left(x\right)\mathbf{+}\mathit{Q}\left(x\right)\mathit{y}\mathbf{+}\mathit{R}\left(x\right){\mathit{y}}^{\mathbf{2}}$is known as Riccati's equation.

(a) A Riccati equation can be solved by a succession of two substitutions provided that we know a particular solution y₁of the equation. Show that the substitution y = y₁ + ureduces Riccati's equation to a Bernoulli equation (4) with n = 2.The Bernoulli equation can then be reduced to a linear equation by the substitution w = u^-1.

(b) Find a one-parameter family of solutions for the differential equation$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}}\mathit{y}\mathbf{+}{\mathit{y}}^{\mathbf{2}}$where y₁ = 2/xis a known solution of the equation.

Short answer

Therefore, the result is

(a)$\frac{dw}{dx}+\left[Q\left(x\right)+2y_{\uparrow }R\left(x\right)\right]w=-R\left(x\right)$

(b)$y=\frac{4x^3}{4c-x^4}+\frac{2}{x}$

Step-by-step solution

Given Information

Let

$y=y_1+u$

Differentiate with respect to x

$\frac{dy}{dx}=\frac{d{y}_1}{dx}+\frac{du}{dx}$

The Riccati differential equation can be written as,

$$\begin{gathered} \frac{d{y}_1}{dx}+\frac{du}{dx}=P\left(x\right)+Q\left(x\right)\left[y_1+u\right]+R\left(x\right){\left[y_1+u\right]}^2 \\ \frac{d{y}_1}{dx}+\frac{du}{dx}=P\left(x\right)+Q\left(x\right)\left[y_1+u\right]+R\left(x\right)\left[y_1^2+2y_{1}u+u^2\right] \\ \frac{d{y}_1}{dx}+\frac{du}{dx}=\left[P\left(x\right)+Q\left(x\right)y_1+R\left(x\right)y_1^2\right]+Q\left(x\right)u+2y_{1}R\left(x\right)u+R\left(x\right)u^2 \end{gathered}$$

Substitute the equation

Substitute $\frac{d{y}_1}{dx}=P\left(x\right)+Q\left(x\right)y_1+R\left(x\right)y_1^2$since y₁ is particular solution of

role="math" localid="1664177303770" $$\begin{gathered} \frac{dy}{dx}=P\left(x\right)+Q\left(x\right)y+R\left(x\right)y^2 \\ \frac{d{y}_1}{dx}+\frac{du}{dx}=\frac{d{y}_1}{dx}+Q\left(x\right)u+2y_{1}R\left(x\right)u+R\left(x\right)u^2 \\ \frac{dy/1}{dx}+\frac{du}{dx}-\frac{dy/1}{dx}=Q\left(x\right)u+2y_{1}R\left(x\right)u+R\left(x\right)u^2 \\ \frac{du}{dx}=\left[Q\left(x\right)u+2y_{1}R\left(x\right)\right]u+R\left(x\right)u^2 \\ \frac{du}{dx}-\left[Q\left(x\right)u+2y_{1}R\left(x\right)\right]u=R\left(x\right)u^2 \end{gathered}$$

So, the equation$y_1\frac{du}{dx}-\left[Q\left(x\right)u+2y_{1}R\left(x\right)\right]u=R\left(x\right)u^2$is Bernouill equation with n = 2.

Multiply by u^-2

$$\begin{gathered} \frac{du}{dx}-\left[Q\left(x\right)u+2y_{1}R\left(x\right)\right]u=R\left(x\right)u^2 \\ {u}^{-2}\frac{du}{dx}-\left[Q\left(x\right)u+2y_{1}R\left(x\right)\right]u^{-1}=R\left(x\right)------>\left(1\right) \end{gathered}$$

Let w = u^-1 Differentiate with respect to x

$\frac{dw}{dx}=-u^{-2}\frac{du}{dx}----->\left(2\right)$

Substitute the equation

Substitute w = u^-1 and (2) in (1)

$$\begin{gathered} -\frac{dw}{dx}-\left[Q\left(x\right)+2y_{1}R\left(x\right)\right]w=R\left(x\right) \\ \frac{dw}{dx}+\left[Q\left(x\right)+2y_{1}R\left(x\right)\right]w=-R\left(x\right) \end{gathered}$$

(a) So, the linear differential equation is$\frac{dw}{dx}+\left[Q\left(x\right)+2y_{1}R\left(x\right)\right]w=-R\left(x\right)$

The Riccati's differential equation is(b)

$\frac{dy}{dx}=P\left(x\right)+Q\left(x\right)y+R\left(x\right)y^2----->\left(4\right)$

Based on (4) and given differential equation, we get

$P\left(x\right)=-\frac{4}{x^2},Q\left(x\right)=-\frac{1}{x},R\left(x\right)=1$

Substitute $y_1=\frac{2}{x},Q\left(x\right)=-\frac{1}{x}andR\left(x\right)=1$in (3).

$$\begin{gathered} \frac{dw}{dx}+\left[-\frac{1}{x}+2.\frac{2}{x}.1\right]w=-1 \\ \frac{dw}{dx}+\left[-\frac{1}{x}+\frac{4}{x}\right]w=-1 \\ \frac{dw}{dx}+\frac{3}{x}w=-1----->\left(5\right) \end{gathered}$$

Linear differential equation of the first order

$w\text{'}+P\left(x\right)w=f\left(x\right)----->\left(6\right)$

Based on (5) and (6), we obtain

$P\left(x\right)=\frac{3}{x}andf\left(x\right)=-1$

Now, substitute the value P(x) and f(x) in the formula

$$\begin{gathered} w\left(x\right)=e^{-\int P\left(x\right)dx}\left[c+\int f\left(x\right)e^{\int P\left(x\right)dx}dx\right] \\ w\left(x\right)=e^{-\int \frac{3}{x}dx}\left[c-\int e^{\int \frac{3}{x}dx}dx\right] \\ w\left(x\right)=e^{-3In\left|x\right|}\left[c-\int e^{3In\left|x\right|}dx\right] \\ w\left(x\right)=e^{In{\left|x\right|}^{-3}}\left[c-\int e^{In{\left|x\right|}^3}dx\right] \\ w\left(x\right)=x^{{}^{-3}}\left[c-\int x^{{}^3}dx\right] \\ w\left(x\right)=x^{{}^{-3}}\left[c-\frac{x^4}{4}\right] \\ w\left(x\right)=\frac{c}{x^3}-\frac{x}{4} \end{gathered}$$

Substitute the value in above equation

Substitute w = u^-1 in the above equation.

$$\begin{gathered} \frac{1}{u}=\frac{c}{x^3}-\frac{x}{4} \\ u=\frac{1}{\frac{c}{x^3}-\frac{x}{4}} \end{gathered}$$

Substitute u = y - y₁ and $y_1=\frac{2}{x}$in the above equation.

$$\begin{gathered} y-\frac{2}{x}=\frac{1}{{\displaystyle \frac{c}{x^3}}-{\displaystyle \frac{x}{4}}} \\ y=\frac{1}{{\displaystyle \frac{c}{x^3}-\frac{x}{4}}}+\frac{2}{x} \\ y=\frac{4x^3}{4c-x^4}+\frac{2}{x} \end{gathered}$$