Question

In problem 7-16, solve the equation.$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{3}{\mathbf{x}}^2{\left(\mathbf{1}\mathbf{+}{\mathbf{y}}^2\right)}^{\frac{3}{2}}$

Short answer

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

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 g(x)that is a function of x alone and h(y) that is a function of y alone.

A separable differential equation can be expressed as role="math" localid="1655920680606" $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{h}\left(y\right)$. 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

$\frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^2{\left(1+{\mathrm{y}}^2\right)}^{\frac{3}{2}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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

$\frac{\mathrm{dy}}{{\left(1+{\mathrm{y}}^2\right)}^{\frac{3}{2}}}=3{\mathrm{x}}^2\mathrm{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

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

$\int \frac{\mathrm{dy}}{{\left(1+{\mathrm{y}}^2\right)}^{\frac{3}{2}}}=\int 3{\mathrm{x}}^2\mathrm{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Evaluate $\int \frac{\mathrm{dy}}{{\left(1+{\mathrm{y}}^2\right)}^{\frac{3}{2}}}$.

Consider $y=\tan z$. The differential is $\mathrm{dy}={\sec }^2\mathrm{z}\mathrm{dz}$.

On substituting, the integration becomes,

role="math" localid="1655921895578" $\begin{array}{c}\int \frac{\mathrm{dy}}{{\left(1+{\mathrm{y}}^2\right)}^{\frac{3}{2}}}=\int \frac{}{{\left(1+{\tan }^2\mathrm{z}\right)}^{\frac{3}{2}}}{\sec }^2\mathrm{z}\mathrm{dz}\\ =\int \frac{1}{{\sec }^3\mathrm{z}}{\sec }^2\mathrm{z}\mathrm{dz}\left[1+{\tan }^2\mathrm{z}={\sec }^2\mathrm{z}\right]\\ =\int \frac{\mathrm{dz}}{\sec \mathrm{z}}\\ =\int \cos \mathrm{z}\mathrm{dz}\end{array}$

role="math" localid="1655921981890" $\begin{array}{c}=\sin \mathrm{z}+\mathrm{C}\left[\mathrm{C}=\mathrm{IntegratingConstant}\right]\\ =\sin \left({\tan }^{-1}\left(\mathrm{y}\right)\right)+\mathrm{C}\\ =\sin \left[{\sin }^{-1}\left(\frac{\mathrm{y}}{\sqrt{1+{\mathrm{y}}^2}}\right)\right]+\mathrm{C}\\ =\frac{\mathrm{y}}{\sqrt{1+{\mathrm{y}}^2}}+\mathrm{C}\end{array}$

Then, equation (3) results,

role="math" localid="1655922136305" $\begin{array}{c}\int \frac{\mathrm{dy}}{{\left(1+{\mathrm{y}}^2\right)}^{\frac{3}{2}}}=\int 3{\mathrm{x}}^2\mathrm{dx}\\ \frac{\mathrm{y}}{\sqrt{1+{\mathrm{y}}^2}}={\mathrm{x}}^3+\mathrm{C}\\ \frac{\mathrm{y}}{\sqrt{1+{\mathrm{y}}^2}}-{\mathrm{x}}^3=\mathrm{C}\end{array}$

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