1 Solved example of first order differential equations $\frac{dy}{dx}=\frac{5x^2}{4y}$ 2 Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$ $4ydy-5x^2dx=0$ 3 The differential equation $4ydy-5x^2dx=0$ is exact, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and they satisfy the test for exactness: $\displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form $f(x,y)=C$ $\frac{dy}{dx}=\frac{5x^2}{4y}$ Intermediate steps Find the derivative of $M(x,y)$ with respect to $y$ $\frac{d}{dy}\left(-5x^2\right)$ The derivative of the constant function ($-5x^2$) is equal to zero 0 Find the derivative of $N(x,y)$ with respect to $x$ $\frac{d}{dx}\left(4y\right)$ The derivative of the constant function ($4y$) is equal to zero 0 4 Using the test for exactness, we check that the differential equation is exact $0=0$ Intermediate steps The integral of a function times a constant ($-5$) is equal to the constant times the integral of the function $-5\int x^2dx$ Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$ $\frac{-5x^{3}}{3}$ Since $y$ is treated as a constant, we add a function of $y$ as constant of integration $\frac{-5x^{3}}{3}+g(y)$ 5 Integrate $M(x,y)$ with respect to $x$ to get $\frac{-5x^{3}}{3}+g(y)$ Intermediate steps The derivative of the constant function ($\frac{-5x^{3}}{3}$) is equal to zero 0 The derivative of $g(y)$ is $g'(y)$ $0+g'(y)$ 6 Now take the partial derivative of $\frac{-5x^{3}}{3}$ with respect to $y$ to get $0+g'(y)$ Intermediate steps Simplify and isolate $g'(y)$ $4y=0+g$ $x+0=x$, where $x$ is any expression $4y=g$ Rearrange the equation $g=4y$ 7 Set $4y$ and $0+g'(y)$ equal to each other and isolate $g'(y)$ $g'(y)=4y$ Intermediate steps Integrate both sides with respect to $y$ $g=\int4ydy$ The integral of a function times a constant ($4$) is equal to the constant times the integral of the function $g=4\int ydy$ Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$ $g=2y^2$ 8 Find $g(y)$ integrating both sides $g(y)=2y^2$ 9 We have found our $f(x,y)$ and it equals $f(x,y)=\frac{-5x^{3}}{3}+2y^2$ 10 Then, the solution to the differential equation is $\frac{-5x^{3}}{3}+2y^2=C_0$ Intermediate steps Combine all terms into a single fraction with $3$ as common denominator $\frac{-5x^{3}+2\cdot 3y^2}{3}=C_0$ Multiply $2$ times $3$ $\frac{-5x^{3}+6y^2}{3}=C_0$ Multiply both sides of the equation by $3$ $-5x^{3}+6y^2=3C_0$ We can rename $3C_0$ as other constant $-5x^{3}+6y^2=C_0$ We need to isolate the dependent variable $y$, we can do that by subtracting $-5x^{3}$ from both sides of the equation $6y^2=5x^{3}+C_0$ Divide both sides of the equation by $6$ $y^2=\frac{5x^{3}+C_0}{6}$ Removing the variable's exponent $y=\pm \sqrt{\frac{5x^{3}+C_0}{6}}$ The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$ $y=\pm \frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}}$ As in the equation we have the sign $\pm$, this produces two identical equations that differ in the sign of the term $\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}}$. We write and solve both equations, one taking the positive sign, and the other taking the negative sign $y=\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}},\:y=\frac{-\sqrt{5x^{3}+C_0}}{\sqrt{6}}$ 11 Find the explicit solution to the differential equation $y=\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}},\:y=\frac{-\sqrt{5x^{3}+C_0}}{\sqrt{6}}$ Final Answer$y=\frac{\sqrt{5x^{3}+C_0}}{\sqrt{6}},\:y=\frac{-\sqrt{5x^{3}+C_0}}{\sqrt{6}}$ What is a first order differential equation with an initial condition?A first order differential equation is an equation of the form F(t,y,y′)=0. F ( t , y , y ′ ) = 0 .
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