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Solve dydx−7y=10\frac{dy}{dx}-7y=10dxdy−7y=10
y=e7x+c7−107y=\frac{e^{7x+c}}{7}-\frac{10}{7}y=7e7x+c−710
y=e7x+c7y=\frac{e^{7x+c}}{7}y=7e7x+c
y=e77−107y=\frac{e^7}{7}-\frac{10}{7}y=7e7−710
Solve the equation
dydx−7yx=x2\frac{dy}{dx}-\frac{7y}{x}=x^2dxdy−x7y=x2
y=−x34+cx7y=-\frac{x^3}{4}+cx^7y=−4x3+cx7
y=x34+cx7y=\frac{x^3}{4}+cx^7y=4x3+cx7
y=x34−cx7y=\frac{x^3}{4}-cx^7y=4x3−cx7
y''+2y'+y=0
y=c1e-t+c2te-t
y=c1et+c2te-t
y=c1et+c2tet
dydx+4yx=x3y2\frac{dy}{dx}+\frac{4y}{x}=x^3y^2dxdy+x4y=x3y2
y=1x4(−lnx+c)y=\frac{1}{x^4\left(-\ln x+c\right)}y=x4(−lnx+c)1
y=1x4(lnx+c)y=\frac{1}{x^4\left(\ln x+c\right)}y=x4(lnx+c)1
y=1(−lnx+c)y=\frac{1}{\left(-\ln x+c\right)}y=(−lnx+c)1
y''+3y'=0
y=c1+c2e-3t
y=c1+c2e3t
y=c1e-3t
y''-2y'+y=0
y=c1e-t+c2tet
What is the general solution to the first-order linear differential equation dy/dx +2y = 3?
y=−e−2x+c2+32y=-\frac{e^{-2x+c}}{2}+\frac{3}{2}y=−2e−2x+c+23
y=−ex+c2+32y=-\frac{e^{x+c}}{2}+\frac{3}{2}y=−2ex+c+23
y=e−2x+c2+32y=\frac{e^{-2x+c}}{2}+\frac{3}{2}y=2e−2x+c+23
dy/dx = 3 cos cos x
y=3sinsinx+cy=3\sin\sin x+cy=3sinsinx+c
y=sinsinx+cy=\sin\sin x+cy=sinsinx+c
y=−3sinsinx+cy=-3\sin\sin x+cy=−3sinsinx+c
dydx=e−y(2x−4)\frac{dy}{dx}=e^{-y}\left(2x-4\right)dxdy=e−y(2x−4)
y=ln(x2−4x+c)y=\ln\left(x^2-4x+c\right)y=ln(x2−4x+c)
y=lnln(x2−4)y=\ln\ln\left(x^2-4\right)y=lnln(x2−4)
y=ln(x2+4x+c)y=\ln\left(x^2+4x+c\right)y=ln(x2+4x+c)
y''+5y'=15x2
y=c2+x3- (3/5) x2+ (6/25) x - (1/5) c1e-5t
y=- (3/5) x2+ (6/25) x - (1/5) c1e-5t
y=c2+x3- (3/5) x2+ (6/25) x + (1/5) c1e-5t
It is done.