First order differential Equations Solutions

First order differential Equations Solutions: First-order differential equations are mathematical tools used to describe relationships between functions and their derivatives. Solving these equations involves employing various techniques, such as separation of variables, the integrating factor method, slope fields, or numerical methods. The solutions provide insights into the behavior of dynamic systems and have wide-ranging applications in physics, engineering, economics, and many other fields.

This Section provides theFirst order differential Equations Solutions by using separation of variables method. Simply Click on the desired question and get step by step solution.

Question # 1: $\frac{d y}{d x}=\frac{x^2}{y\left(1+x^3\right)}$

Check Step by Step Solution

$\begin{aligned}& \frac{d y}{d x}=\frac{x^2}{y\left(1+x^3\right)} \\& y d y=\frac{x^2}{1+x^3} d x \\& \int y d y=\frac{1}{3} \int \frac{3 x^2}{1+x^3} \cdot d x \\& \frac{y^2}{2}=\frac{1}{3} \ln \left(1+x^3\right)+c \\& \frac{3 y^2}{2}=\ln \left(1+x^3\right)+3 c \\& 3 y^2=2 \ln \left(1+x^3\right)+6 c \\& 3 y^{2}=2 \ln \left(1+x^3\right)+c^{\prime} \\&\end{aligned}$

Question # 2: $\frac{d y}{d x}+y^{2} \sin x=0$

Check Step by Step Solution

Given equation is

$\frac{d y}{d x}+y^{2} \sin x=0$

$\Rightarrow \frac{d y}{d x}=-y^{2} \sin x$

$\Rightarrow \frac{d y}{y^{2}}=-\sin x d x$

Integrating both sides

$\int \frac{d y}{y^{2}}=-\int \sin x d x$

$\Rightarrow \frac{-1}{y}=-(-\cos x)+c$

$\Rightarrow \frac{1}{\mathrm{y}}+\cos \mathrm{x}=\mathrm{c}$

is required solution.

Question #3: $\frac{d y}{d x}=1+x+y^{2}+x y^{2}$

Check Step by Step Solution

Given equation is

$\frac{d y}{d x}=1+x+y^{2}+x y^{2}$

$\Rightarrow \frac{d y}{d x}=1(1+x)+y^{2}(1+x)$

$\Rightarrow \frac{d y}{d x}=\left(1+y^{2}\right)(1+\mathrm{x})$

$\Rightarrow \frac{d y}{1+y^{2}}=(1+\mathrm{x}) \mathrm{dx}$

Integrating both sides

$\Rightarrow \int \frac{\mathrm{dy}}{1+y^{2}}=\int(1+x) d x$

$\Rightarrow \tan ^{-1}(y)=\int d x+\int x d x$

$\Rightarrow \tan ^{-1}(\mathrm{y})-\mathrm{x}-\frac{\mathrm{x}^{2}}{2}=\mathrm{c}$

is required solution.

Question # 4: $(x y+2 x+y+2) d x+\left(x^{2}+2 x\right) d y=0$

Check Step by Step Solution

Given equation is

$(x y+2 x+y+2) d x+\left(x^{2}+2 x\right) d y=0$

$\Rightarrow(x y+2 x+y+2) d x=-\left(x^{2}+2 x\right) d y$

$\Rightarrow \mathrm{y}(\mathrm{x}+1)+2(\mathrm{x}+1) \mathrm{dx}=-\left(\mathrm{x}^{2}+2 \mathrm{x}\right) \mathrm{dy}$

$\Rightarrow(\mathrm{y}+2)(\mathrm{x}+1) \mathrm{dx}=-\left(\mathrm{x}^{2}+2 \mathrm{x}\right) \mathrm{dy}$

$\Rightarrow \frac{d y}{y+2}=-\frac{x+1}{\left(x^{2}+2 x\right)} d x$

Integrating both sides

$\int \frac{d y}{y+2}=-\int \frac{x+1}{\left(x^{2}+2 x\right)} d x$

$\Rightarrow \int \frac{d y}{y+2}=-\frac{1}{2} \int \frac{2 x+2}{x^{2}+2 x} d x$ $\Rightarrow \ln (\mathrm{y}+2)=-\frac{1}{2} \ln \left(\mathrm{x}^{2}+2 \mathrm{x}\right)+\mathrm{c}$

is required solution.

Question #5: $\frac{d y}{d x}=2 x^{2}+y+x^{2} y+x y-2 x-2$

Check Step by Step Solution

Given equation is

$\frac{d y}{d x}=2 x^{2}+y+x^{2} y+x y-2 x-2$

$\Rightarrow \frac{d y}{d x}=y\left(1-x^{2}+x\right)+2\left(x^{2}-x-1\right)$

$\Rightarrow \frac{d y}{d x}=y\left(1-x^{2}+x\right)-2\left(1-x^{2}+x\right)$

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

$\Rightarrow \frac{d y}{y-2}=\left(1-x^{2}+x\right) d x$

Integrating both sides

$\Rightarrow \int \frac{d y}{y-2}=\int\left(1-x^{2}+x\right) d x$

$\Rightarrow \int \frac{d y}{y-2}=\int d x-\int x^{2} d x+\int x d x$

$\Rightarrow \ln (\mathrm{y}-2)=\mathrm{x}-\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{x}^{2}}{2}+\mathrm{c}_{1}$

$\Rightarrow 6 \ln (\mathrm{y}-2)=6 \mathrm{x}-2 \mathrm{x}^{3}+3 \mathrm{x}^{2}+6 \mathrm{c}_{1}$

$\Rightarrow-6 \ln (\mathrm{y}-2)=-6 x+2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-6 \mathrm{c}_{1}$

$\Rightarrow \ln (\mathrm{y}-2)^{-6}=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-6 \mathrm{x}+\mathrm{c}$

$\Rightarrow(\mathrm{y}-2)^{-6}=\mathrm{e}^{2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-6 \mathrm{x}+\mathrm{c}}$

is required solution.

Question # 6: $\operatorname{cosec} y d x+\sec x d y=0$

Check Step by Step Solution

Given equation is

$\operatorname{cosec} y d x+\sec x d y=0$

$\Rightarrow \operatorname{cosecydx}=-\operatorname{cosecy} d x$

$\Rightarrow \frac{d x}{\sec x}=-\frac{d y}{\operatorname{cosec} y}$

Integrating both sides

$\Rightarrow \int \frac{\mathrm{dx}}{\sec x}=-\int \frac{\mathrm{dy}}{\operatorname{cosecy}}$

$\Rightarrow \int \cos x d x=-\int \sin y d y$

$\Rightarrow \sin \mathrm{x}=-(-\cos \mathrm{x})$

$\Rightarrow \sin x=\cos x+c$

$\Rightarrow \sin \mathrm{x}-\cos \mathrm{x}=\mathrm{c}$

is required solution.

Question # 7: $y(1+x) d x+x(1+y) d y$

Check Step by Step Solution

Given equation is

$y(1+x) d x+x(1+y) d y$

$\Rightarrow \mathrm{y}(1+\mathrm{x}) \mathrm{d} x=-\mathrm{x}(1+\mathrm{y}) \mathrm{dy}$

$\Rightarrow \frac{(1+x)}{x} d x=-\frac{(1+y)}{y} d y$

Integrating both sides

$\Rightarrow \int \frac{(1+\mathrm{x})}{\mathrm{x}} \mathrm{dx}=-\int \frac{(1+\mathrm{y})}{\mathrm{y}} \mathrm{dy}$

$\Rightarrow \int \frac{d x}{x}+\int \frac{x}{x} d x=-\int \frac{d y}{y}-\int \frac{y}{y} d y$

$\Rightarrow \ln \mathrm{x}+\mathrm{x}=-\ln y-\mathrm{y}+\mathrm{c}$

$\Rightarrow \mathrm{x}+\mathrm{y}+\ln \mathrm{x}+\ln \mathrm{y}=\mathrm{c}$

$\Rightarrow \mathrm{x}+\mathrm{y}+\ln |\mathrm{xy}|=\mathrm{c}$

is required solution.

Question # 8: $y \sqrt{1+x^{2}} d x+x \sqrt{1+y^{2}} d y=0$

Check Step by Step Solution

Given equation is

$y \sqrt{1+x^{2}} d x+x \sqrt{1+y^{2}} d y=0$

$\Rightarrow \mathrm{y} \sqrt{1+\mathrm{x}^{2}} \mathrm{dx}=-\mathrm{x} \sqrt{1+\mathrm{y}^{2}} \mathrm{dy}$

$\Rightarrow \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}} \mathrm{dx}=-\frac{\sqrt{1+\mathrm{y}^{2}}}{\mathrm{y}} \mathrm{dy}$

Integrating both sides

$\Rightarrow \int \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}} \mathrm{dx}=-\int \frac{\sqrt{1+\mathrm{y}^{2}}}{\mathrm{y}} \mathrm{dy}---$

Consider $\mathrm{I}_{1}=\int \frac{\sqrt{1+\mathrm{x}^{2}}}{\mathrm{x}} \mathrm{dx}$

$\begin{gathered}\text { put } x=\tan \theta \\dx=\sec ^{2} \theta\end{gathered}$

Therefore,

$\begin{aligned}&I_{1}<em>=\int \frac{\sec \theta \cdot \sec ^{2} \theta}{\tan \theta} \\ & I_{1}</em>=\int \frac{\sec \theta \cdot\left(1+\tan ^{2} \theta\right) \mathrm{d} \theta}{\tan \theta} \\& I_{1}=\int \frac{\sec \theta}{\tan \theta} \mathrm{d} \theta+\int \frac{\sec \theta \cdot \tan ^{2} \theta}{\tan \theta} \mathrm{d} \theta \\& I_{1}=\int \frac{1 / \cos \theta}{\sin \theta / \cos \theta} d \theta+\int \sec \theta \cdot \tan \theta d \theta \\& I_{1}<em>=\int \frac{1}{\sin \theta} \mathrm{d} \theta+\int \sec \theta \cdot \tan \theta \mathrm{d} \theta \\ & </em>I_{1}=\int \operatorname{cosec} \theta \mathrm{d} \theta+\int \sec \theta \cdot \tan \theta \mathrm{d} \theta \\& I_{1}=\ln (\operatorname{cosec} \theta-\cot \theta)+\sec \theta \\& x=\tan \theta \\& \cot \theta=\frac{1}{\mathrm{x}} \\& \operatorname{cosec} \theta=\sqrt{\cot ^{2} \theta+1} \\& \operatorname{cosec} \theta=\sqrt{\frac{1}{\mathrm{x}^{2}}+1} \\& \operatorname{cosec} \theta=\frac{\sqrt{1+x^{2}}}{x}\end{aligned}$

Therefore,

$I_{1}=\ln \left(\frac{\sqrt{1+x^{2}}}{x}-\frac{1}{x}\right)+\sec \theta$

$\Rightarrow \int \frac{\sqrt{1+x^{2}}}{x} d x=\ln \left(\frac{\sqrt{1+x^{2}}-1}{x}\right)+\sqrt{1+x^{2}}$

Similarly,

$\int \frac{\sqrt{1+y^{2}}}{\mathrm{y}} d y=\ln \left(\frac{\sqrt{1+y^{2}}-1}{\mathrm{y}}\right)+\sqrt{1+y^{2}}$

Therefore, (1) becomes

$\begin{aligned}& \ln \left(\frac{\sqrt{1+x^{2}}-1}{x}\right)+\sqrt{1+x^{2}} \& =-\ln \left(\frac{\sqrt{1+\mathrm{y}^{2}}-1}{\mathrm{y}}\right)-\sqrt{1+\mathrm{y}^{2}} \& +\mathrm{c} \& \Rightarrow \sqrt{1+\mathrm{x}^{2}}+\sqrt{1+\mathrm{y}^{2}} \& +\ln \left(\frac{\left(\sqrt{1+x^{2}}-1\right)\left(\sqrt{1+y^{2}}-1\right)}{x y}\right) \& =\mathrm{c}\end{aligned}$

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Differential Equation Solutions