Daniell Phillips

2021-12-26

First-Order Linear Differential Equations; Solutions Suggested by the Equation
$\frac{dy}{dx}={\left(x+y+1\right)}^{2}$

Hector Roberts

Expert

Step 1
We have given the differential equation as $\frac{dy}{dx}={\left(x+y+1\right)}^{2}$. To solve this differential equation, we have to use substitution to convert the equation into the separable differential equation because separable equations are easy to solve.
Step 2
So substitute . First, differentiate $x+y+1=v$ with respect to x to find the expression for $\frac{dy}{dx}$.
$\frac{d}{dx}\left(x+y+1\right)=v$
$1+\frac{dy}{dx}=\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{dv}{dx}-1$
Now, do substitution and simplify the differential equation.
$\frac{dv}{dx}-1={v}^{2}$
$\frac{dv}{dx}=1+{v}^{2}$
$\frac{dv}{1+{v}^{2}}=dx$
Step 3
Now, integrate both sides of $\frac{dv}{1+{v}^{2}}=dx$ and evaluate the integral. Use the formula $\int \frac{du}{1+{u}^{2}}={\mathrm{tan}}^{-1}u+C$.
$\int \frac{dv}{1+{v}^{2}}=\int dx$
${\mathrm{tan}}^{-1}v=x+C$
${\mathrm{tan}}^{-1}\left(x+y+1\right)=x+C$ [Substitute back $x+y+1=v$]
Hence, the solution of the given differential equation is ${\mathrm{tan}}^{-1}\left(x+y+1\right)=x+C$, where C is an integral constant.

sirpsta3u

Expert

$\frac{dy}{dx}={\left(x+y+1\right)}^{2}$ (1)
Let. $x+y+1=t$
$1+\frac{dy}{dx}=\frac{dt}{dx}$
$\frac{dy}{dx}=\frac{dt}{dx}-1$ (2)
from eqn(1) and (2)
$\frac{dt}{dx}-1={t}^{2}$
$\frac{dt}{dx}=1+{t}^{2}$
$1.\frac{dt}{1+{t}^{2}}=dx$
${\mathrm{tan}}^{-1}\left(t\right)=x+C$
$t=\mathrm{tan}\left(x+C\right)$
$x+y+1=\mathrm{tan}\left(x+C\right)$
$y=\mathrm{tan}\left(x+C\right)-x-1$. Answer.

Vasquez

Expert

The given differential equation is
$\frac{dy}{dx}=\left(x+y+1{\right)}^{2}$ ...(i)
Let $x+y+1=v$
$⇒\frac{dy}{dx}=\frac{dv}{dx}-1$
$\frac{dv}{dx}-1={v}^{2}$
$⇒\frac{dv}{{v}^{2}+1}=dx$
Integrating, we get
$\int \frac{dv}{{v}^{2}+1}=\int dx$
${\mathrm{tan}}^{-1}\left(v\right)=x+c$
${\mathrm{tan}}^{-1}\left(x+y+1\right)=x+c$
Which is the required solution.

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