Solve the differential equation x^3dx+y^0.5dy=0

x^3dx+\sqrt{y}dy=0

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Answer

$y=NaN\sqrt[3]{x^{8}}+C_0$

Step by step solution

Problem

$x^3dx+\sqrt{y}dy=0$
1

Grouping the terms of the differential equation

$\sqrt{y}dy=-x^3dx$
2

Integrate both sides, the left side with respect to $y$, and the right side with respect to $x$

$\int\sqrt{y}dy=\int-x^3dx$
3

Taking the constant out of the integral

$\int\sqrt{y}dy=-\int x^3dx$
4

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{2}{3}\sqrt{y^{3}}=-\int x^3dx$
5

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a constant function

$\frac{2}{3}\sqrt{y^{3}}=-\frac{x^{4}}{4}$
6

Multiply both sides of the equation by $$

$\sqrt{y^{3}}=-\frac{3}{2}\cdot\frac{x^{4}}{4}$
7

Simplify the fraction

$\sqrt{y^{3}}=-\frac{3}{8}x^{4}$
8

Removing the variable's exponent

$y^{div\left(1,\frac{3}{2}\right)\cdot \frac{3}{2}}=\left(-\frac{3}{8}x^{4}\right)^{div\left(1,\frac{3}{2}\right)}$
9

Divide $1$ by $\frac{3}{2}$

$y^{\frac{2}{3}\cdot \frac{3}{2}}=\sqrt[3]{\left(-\frac{3}{8}x^{4}\right)^{2}}$
10

Multiply $\frac{3}{2}$ times $\frac{2}{3}$

$y^{1}=\sqrt[3]{\left(-\frac{3}{8}x^{4}\right)^{2}}$
11

Any expression to the power of $1$ is equal to that same expression

$y=\sqrt[3]{\left(-\frac{3}{8}x^{4}\right)^{2}}$
12

The power of a product is equal to the product of it's factors raised to the same power

$y=NaN\sqrt[3]{x^{8}}$
13

Add the constant of integration

$y=NaN\sqrt[3]{x^{8}}+C_0$

Answer

$y=NaN\sqrt[3]{x^{8}}+C_0$

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Problem Analysis

Main topic:

First order differential equations

Time to solve it:

0.26 seconds

Views:

104