Question

In problem 7-16, solve the equation.$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{x}}{{\mathbf{y}}^{\mathbf{2}}\sqrt{\mathbf{1}\mathbf{+}\mathbf{x}}}$

Short answer

The solution of the given differential equation is${\mathbf{y}}^3\mathbf{=}\mathbf{2}\left[\mathbf{3}\mathbf{x}\sqrt{1+x}\mathbf{-}\mathbf{2}{\left(1+x\right)}^{\frac{3}{2}}\right]\mathbf{+}\mathbf{C}$.

Step-by-step solution

Concept of Separable Differential Equation

A first-order ordinary differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions $\mathbf{g}\left(x\right)$that is a function of$\mathbf{x}$alone and$\mathbf{h}\left(y\right)$that is a function of$\mathbf{y}$alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{h}\left(y\right)$. By separating the variables, the equation follows $\frac{\mathrm{dy}}{\mathbf{h}\left(y\right)}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{dx}$. Then, on direct integration of both sides, the solution of the differential equation is determined.

Solution of the Equation

The given equation is

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{{\mathrm{y}}^2\sqrt{1+\mathrm{x}}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

After separating the variables, equation (1) can be written as

${\mathrm{y}}^2\mathrm{dy}=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}}}\mathrm{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Integrate both sides of equation (2). It results,

$\begin{array}{c}\int {\mathrm{y}}^2\mathrm{dy}=\int \frac{\mathrm{x}}{\sqrt{1+\mathrm{x}}}\mathrm{dx}\\ \int {\mathrm{y}}^2\mathrm{dy}=2\int \frac{\mathrm{x}}{2\sqrt{1+\mathrm{x}}}\mathrm{dx}\\ \int {\mathrm{y}}^2\mathrm{dy}=2\int \mathrm{x}\mathrm{d}\left(\sqrt{1+\mathrm{x}}\right)\\ \int {\mathrm{y}}^2\mathrm{dy}=2\left[\mathrm{x}\sqrt{1+\mathrm{x}}-\int \sqrt{1+\mathrm{x}}\mathrm{dx}\right]\left[\int \mathrm{u}\mathrm{dv}=\mathrm{uv}-\int \mathrm{v}\mathrm{du}\right]\end{array}$

$\begin{array}{c}\frac{{\mathrm{y}}^3}{3}=2\left[\mathrm{x}\sqrt{1+\mathrm{x}}-\frac{2}{3}{\left(1+\mathrm{x}\right)}^{\frac{3}{2}}\right]+\mathrm{k}\left[\mathrm{k}=\mathrm{IntegratingConstant}\right]\\ {\mathrm{y}}^3=2\left[3\mathrm{x}\sqrt{1+\mathrm{x}}-2{\left(1+\mathrm{x}\right)}^{\frac{3}{2}}\right]+3\mathrm{k}\\ {\mathrm{y}}^3=2\left[3\mathrm{x}\sqrt{1+\mathrm{x}}-2{\left(1+\mathrm{x}\right)}^{\frac{3}{2}}\right]+\mathrm{C}\left[\mathrm{C}=3\mathrm{k}=\mathrm{Constant}\right]\end{array}$

Therefore, the solution of the given equation is ${\mathbf{y}}^3\mathbf{=}\mathbf{2}\left[\mathbf{3}\mathbf{x}\sqrt{1+x}\mathbf{-}\mathbf{2}{\left(1+x\right)}^{\frac{3}{2}}\right]\mathbf{+}\mathbf{C}$.