Separable Differential Equations

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Separable (First Order) Differential Equations
Key Identifiers
Separable first order DEs can be put in the form dydx=f(x)⋅g(y)\frac{dy}{dx}=f\left(x\right)\cdot g\left(y\right)dxdy​=f(x)⋅g(y)

Strategies for Solving Separable DEs

IVP
If an initial condition is given, use it to solve for the unknown constant C.

Practice Question
Solve the DE dPdt=tt2−1PeP\frac{dP}{dt}=\frac{t\sqrt{t^2-1}}{Pe^P}dtdP​=PePtt2−1​​.

P=ln⁡[(t2−1)3t]+CP=\ln\left[\frac{\sqrt{\left(t^2-1\right)^3}}{t}\right]+CP=ln[t(t2−1)3​​]+C

P=(t2−1)33et+CP=\frac{\sqrt{\left(t^2-1\right)^3}}{3e^t}+CP=3et(t2−1)3​​+C

P=(t2−1)33et+1+CP=\frac{\sqrt{\left(t^2-1\right)^3}}{3e^t}+1+CP=3et(t2−1)3​​+1+C

ePP=13(t2−1)3+1+Ce^PP=\frac{1}{3}\sqrt{\left(t^2-1\right)^3}+1+CePP=31​(t2−1)3​+1+C

eP(P−1)=13(t2−1)3+Ce^P\left(P-1\right)=\frac{1}{3}\sqrt{\left(t^2-1\right)^3}+CeP(P−1)=31​(t2−1)3​+C

I don't know

Practice Question
Suppose that we know x2y′=yx^2y' = y x2y′=yand y(1)=1y(1)=1y(1)=1. Determine the equation of yyy. 

Practice Question
Given that dydx=4x3(y−1)\frac{dy}{dx}=4x^3\left(y-1\right)dxdy​=4x3(y−1) and y(0)=2y\left(0\right)=2y(0)=2, find y(1)y\left(1\right)y(1).

Practice Question
Given that dydx=ey−3x\frac{dy}{dx}=e^{y-3x}dxdy​=ey−3x and y(0)=ln⁡3y\left(0\right)=\ln3y(0)=ln3, find y(ln⁡2)y\left(\ln2\right)y(ln2).

Hint: you will need to simplify your answer using ln⁡\lnln rules
r ln⁡x=ln⁡(xr)r\ \ln x=\ln\left(x^r\right)r lnx=ln(xr)
eln⁡x=xe^{\ln x}=xelnx=x

Practice Question
Solve dydx=4e5xy\frac{dy}{dx}=\frac{4e^{5x}}{y}dxdy​=y4e5x​

y=45e5x+Cy=\frac{4}{5}e^{5x}+Cy=54​e5x+C

y=85e5x+Cy=\frac{8}{5}e^{5x}+Cy=58​e5x+C

y=±25e5x+Cy=\pm\sqrt{\frac{2}{5}e^{5x}+C}y=±52​e5x+C​

y=±45e5x+Cy=\pm\sqrt{\frac{4}{5}e^{5x}+C}y=±54​e5x+C​

y=±85e5x+Cy=\pm\sqrt{\frac{8}{5}e^{5x}+C}y=±58​e5x+C​

I don't know

Wize University Calculus 2 Textbook > Differential Equations

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Separable (First Order) Differential Equations

Key Identifiers

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IVP

Practice Question

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