Differential Equations - First-Order Differential Equations

Solve the following differential equation

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

$y=Ce^x^{^{3}}$

$y=e^x^{^{3}}$

$y=e^x^{^{3}+C}$

$ln(y)=e^x^{^{3}}$

Explanation

So this is a separable differential equation. The first step is to move all of the x terms (including dx) to one side, and all of the y terms (including dy) to the other side.

So the differential equation we are given is: $\frac{dy}{dx}=3x^2y$

Which rearranged looks like: $\frac{dy}{y}=3x^2dx$

At this point, in order to solve for y, we need to take the anti-derivative of both sides: $\int \frac{dy}{y}=\int 3x^2dx$

Which equals:

And since this an anti-derivative with no bounds, we need to include the general constant C

So, solving for y, we raise e to the power of both sides:

$e^{ln(y)}=e^{x^3+C}$ which, simplified gives us our answer:

$y=Ce^{x^3}$

Find the general solution of the differential equation

$\frac{dy}{dx} - 5y = -\frac{5}{2}xy^3$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)$

$y = \left( x - \frac{1}{10} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} + Ce^{-10x} \right)^{-1/2}$

None of the other answers

Explanation

This is a Bernoulli Equation of the form

$\frac{dy}{dx} + P(x) = Q(x)y^n$

which requires a substitution

$v = y^{1-n}$

to transform it into a linear equation

Rearranging our equation gives us

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

Substituting $v = y^{-2}$

$-\frac{1}{2} \frac{dv}{dx} - 5v = -\frac{5}{2}x$

$\frac{dv}{dx} + 10v = 5x$

Solving the linear ODE gives us

$v = \frac{x}{2} - \frac{1}{20} + Ce^{-10x}$

Substituting in $v = y^{-2}$ and solving for

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

Find the solution for the following differential equation:

where .

$y = \frac{e^{t^3} - 1}{3}$

$y = -\frac{1}{3}$

$y = \frac{e^{t^3}}{3}$

$y = e^{t^3}-\frac{1}{3}$

Explanation

This equation can be put into the form as follows:

. Differential equations in this form can be solved by use of integrating factor. To solve, take and solve for $\mu = e^{\int p(t)dt} = e^{\int -3t^2dt} = e^{-t^3}$

Note, when using integrating factors, the +C constant is irrelevant as we only need one solution, not infinitely many. Thus, we have set C to 0.

Next, note that $(\mu y)' = y'\mu + \mu'y = e^{-t^3}(y' - 3t^2y) = \mu q(t) = e^{-t^3}t^2$

Or more simply, $(\mu y)' = t^2e^{-t^3}$ . Integrating both sides using substitution of variables we find

$\mu y = \int t^2e^{-t^3}dt = -\frac{e^{-t^3}}{3} + C$

Finally dividing by $\mu = e^{-t^3}$ , we see

$y = -\frac{1}{3} + Ce^{t^3}$ . Plugging in our initial condition,

$0 = -\frac{1}{3} + C$

So $C = \frac{1}{3}$

And $y = \frac{e^{t^3} - 1}{3}$ .

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

$x\frac{dy}{dx}+3y=x^4-x$

$\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^4-\frac{1}{4}x^2+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^3-\frac{1}{4}x+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^4-\frac{1}{4}x^3+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\text{Transient Term}=cx^{3}$

Explanation

First, divide by on both sides of the equation.

$\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\frac{dy}{dx}+\frac{3y}{x}=x^3-1$

Identify the factor term.

$p(x)=\frac{3}{x}$

Integrate the factor.

$e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3$

Substitute this value back in and integrate the equation.

$\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c$

Now divide by to get the general solution.

$\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}$

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is $cx^{-3}$ .

Solve the general solution for the ODE:

$\frac{dy}{dx}=x^8y$

$\ln(y) = \frac{x^9}{9}+C$

where C is an arbitrary constant

$y = \frac{x^9}{9}+C$

where C is an arbitrary constant

$\ln(y) = x^8+C$

where C is an arbitrary constant

$\ln(y) = e^\frac{x^9}{9}+C$

where C is an arbitrary constant

Explanation

First the differential equation can be separated to:

$\frac{dy}{y} = x^8dx$

And then integrated simply to:

$\ln(y) = \frac{x^9}{9}+C$

Find the general solution of the differential equation

$\frac{dy}{dx} - 5y = -\frac{5}{2}xy^3$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)$

$y = \left( x - \frac{1}{10} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} + Ce^{-10x} \right)^{-1/2}$

None of the other answers

Explanation

This is a Bernoulli Equation of the form

$\frac{dy}{dx} + P(x) = Q(x)y^n$

which requires a substitution

$v = y^{1-n}$

to transform it into a linear equation

Rearranging our equation gives us

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

Substituting $v = y^{-2}$

$-\frac{1}{2} \frac{dv}{dx} - 5v = -\frac{5}{2}x$

$\frac{dv}{dx} + 10v = 5x$

Solving the linear ODE gives us

$v = \frac{x}{2} - \frac{1}{20} + Ce^{-10x}$

Substituting in $v = y^{-2}$ and solving for

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

$x\frac{dy}{dx}+3y=x^4-x$

$\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^4-\frac{1}{4}x^2+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^3-\frac{1}{4}x+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^4-\frac{1}{4}x^3+cx^{-3} \\\text{Transient Term}=cx^{-3}$

$\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\text{Transient Term}=cx^{3}$

Explanation

First, divide by on both sides of the equation.

$\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\frac{dy}{dx}+\frac{3y}{x}=x^3-1$

Identify the factor term.

$p(x)=\frac{3}{x}$

Integrate the factor.

$e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3$

Substitute this value back in and integrate the equation.

$\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\x^3y=\int x^6-x^3dx \\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c$

Now divide by to get the general solution.

$\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}$

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is $cx^{-3}$ .

Find the solution for the following differential equation:

where .

$y = \frac{e^{t^3} - 1}{3}$

$y = -\frac{1}{3}$

$y = \frac{e^{t^3}}{3}$

$y = e^{t^3}-\frac{1}{3}$

Explanation

This equation can be put into the form as follows:

. Differential equations in this form can be solved by use of integrating factor. To solve, take and solve for $\mu = e^{\int p(t)dt} = e^{\int -3t^2dt} = e^{-t^3}$

Note, when using integrating factors, the +C constant is irrelevant as we only need one solution, not infinitely many. Thus, we have set C to 0.

Next, note that $(\mu y)' = y'\mu + \mu'y = e^{-t^3}(y' - 3t^2y) = \mu q(t) = e^{-t^3}t^2$

Or more simply, $(\mu y)' = t^2e^{-t^3}$ . Integrating both sides using substitution of variables we find

$\mu y = \int t^2e^{-t^3}dt = -\frac{e^{-t^3}}{3} + C$

Finally dividing by $\mu = e^{-t^3}$ , we see

$y = -\frac{1}{3} + Ce^{t^3}$ . Plugging in our initial condition,

$0 = -\frac{1}{3} + C$

So $C = \frac{1}{3}$

And $y = \frac{e^{t^3} - 1}{3}$ .

Solve the general solution for the ODE:

$\frac{dy}{dx}=x^8y$

$\ln(y) = \frac{x^9}{9}+C$

where C is an arbitrary constant

$y = \frac{x^9}{9}+C$

where C is an arbitrary constant

$\ln(y) = x^8+C$

where C is an arbitrary constant

$\ln(y) = e^\frac{x^9}{9}+C$

where C is an arbitrary constant

Explanation

First the differential equation can be separated to:

$\frac{dy}{y} = x^8dx$

And then integrated simply to:

$\ln(y) = \frac{x^9}{9}+C$

Solve the following differential equation

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

$y=Ce^x^{^{3}}$

$y=e^x^{^{3}}$

$y=e^x^{^{3}+C}$

$ln(y)=e^x^{^{3}}$

Explanation

So this is a separable differential equation. The first step is to move all of the x terms (including dx) to one side, and all of the y terms (including dy) to the other side.

So the differential equation we are given is: $\frac{dy}{dx}=3x^2y$

Which rearranged looks like: $\frac{dy}{y}=3x^2dx$

At this point, in order to solve for y, we need to take the anti-derivative of both sides: $\int \frac{dy}{y}=\int 3x^2dx$

Which equals:

And since this an anti-derivative with no bounds, we need to include the general constant C

So, solving for y, we raise e to the power of both sides:

$e^{ln(y)}=e^{x^3+C}$ which, simplified gives us our answer:

$y=Ce^{x^3}$

First-Order Differential Equations

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation