How to Solve Linear Differential Equations

Ans. A linear differential equation is a linear equation or polynomial with one or more terms consisting of the derivatives of the dependent variable with respect to one or more independent variables.

Ans. A linear equation will always have a solution for all x and y values, but nonlinear equations may or may not have a solution for all x and y values.

Ans. Linear differential equations have constant coefficients and involve the unknown function and its derivatives in a linear way. Nonlinear differential equations do not have these properties, making them more complex to solve.

Ans. Given, The linear differential equation = \(\frac{d y}{d x}+3 y=e^{-2 x}\)

Now, comparing this with the linear differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = 3 and Q = -2x

Hence, we have the Integration factor as I.F which is = \(e^{\int p.d x}\)

\(=e^{\int 3.d x}\)

Further, the solution of the given differential equation will be as follows:

\(y(I . F)=\int Q(I . F) d x+C\)

\(y e^{3 x}=\int e^{-2 x} e^{3 x} d x+C\)

\(y e^{3 x}=\int e^{x} d x+C\)

\(y e^{3 x}=e x+C\)

Now, dividing e^3x on both sides, we get:

\(y=\frac{e^{x}}{e^{3 x}}+\frac{C}{e^{3 x}}\)

\(y=e^{-2 x}+C e^{-3 x}\)

Therefore, the solution is: \(y=e^{-2 x}+C e^{-3 x}\)

Ans. Given, The linear differential equation = \(\frac{d y}{d x}+\frac{y}{x}=x^{2}\)

Now, comparing this with the linear differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = \(\frac{1}{x}\) and Q = x²

Hence, we have the Integration factor as I.F which is = \(e^{\int p.d x}\)

\(=e^{\int \frac{1}{x}.d x}\)

\(=e^{\log x}\)

\(=x\)

Further, the solution of the given differential equation will be as follows:

\(y(I . F)=\int Q(I . F) d x+C\)

\(y x=\int x^{2} \cdot x \cdot d x+C\)

\(xy=\frac{x^4}{4}+ C\)

Now, dividing by x on both sides, we get:

\(y=\frac{x^3}{4}+ C\)

Therefore, the solution is: \(y=\frac{x^3}{4}+ C\)

Ans. Given, the differential equation = \(x \frac{d y}{d x}+2 y=x^{2}\)

This equation can now be simplified to represent the following linear differential equation.

\(\frac{d y}{d x}+y\cdot \frac{2}{x}=x\)

Now, comparing this with the linear differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = \(\frac{2}{x}\) and Q = x

Hence, we have the Integration factor as I.F which is = \(e^{\int p.d x}\)

\(=e^{\int \frac{2}{x}.d x}\)

\(=e^{2\log x}\)

\(=(e^{\log x})^2\)

\(=x^2\)

Further, the solution of the given differential equation will be as follows:

\(y(I . F)=\int Q(I . F) d x+C\)

\(y \cdot x^{2}=\int x \cdot x^{2} \cdot d x+C \)

\(y \cdot x^{2}=\int x^{3} \cdot d x+C \)

\(y \cdot x^{2}=\frac{x^{4}}{4}+C\)

\(x^{4}-4 x^{2} y+C=0\)

Therefore, the solution is: \(x^{4}-4 x^{2} y+C=0\)

Ans. Given, the LDE = \(\frac{d y}{d x}=\left[\frac{1}{1+x^{3}}\right]-\left[\frac{3 x^{2}}{\left(1+x^{2}\right)}\right] y\)

This equation can be rewritten as:

\(\frac{d y}{d x}+\left[\frac{3 x^{2}}{\left(1+x^{3}\right)}\right] y=\frac{1}{\left(1+x^{3}\right)}\)

Now, comparing this with the linear differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = \(\frac{3x^2}{1+x^3}\) and Q = \(\frac{1}{1+x^3}\)

Hence, we have the Integration factor as I.F which is = \(e^{\int p.d x}\)

\(=e^{\int {\frac{3 x^{2}}{1+x^{3}}}} d x\)

\(=e^{\int\left(1+x^{3}\right)}\)

\(=1+x^{3}\)

Again here, we can rewrite the L.H.S as: \(\frac{d(y.I.F)}{dx}\)

\(d\left[y \cdot\left(1+x^{3}\right)\right] d x=\left[\frac{1}{1+x^{3}}\right] \cdot\left(1+x^{3}\right)\)

Further, integrating both the sides with respect to x , we get:

\(y .(1+x^3)= x\)

\(y=\frac{x}{\left(1+x^{3}\right)}+C\)

Therefore, the solution is: \(y=\frac{x}{\left(1+x^{3}\right)}+C\)

Ans. Given, the LDE = \(\cos ^{2} x \cdot \frac{d y}{d x}+y=\tan x\)

The equation can be rewritten as: 

\(\frac{d y}{d x}+\frac{y}{\cos ^{2} x}=\tan x / \cos ^{2} x\)

\(\frac{d y}{d x}+\left(\sec ^{2} x\right) y=\tan x \cdot \sec ^{2} x\)

Now, comparing this with the linear differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = sec²x and Q = tan x.sec²x

Hence, we have the Integration factor as I.F which is = \(e^{\int p.d x}\)

\(=e^{\int sec^2x.d x}\)

\(=e^{\int \tan x}\)

Further, the solution of the given differential equation will be as follows:

\(y(I . F)=\int Q(I . F) d x+C\)

\(y \cdot e^{\tan x}=\int \tan x \cdot \sec ^{2} x \cdot e^{\tan x} d x+C\)

Now, let’s put tan x = t

\(\sec ^{2} x \frac{d x}{d t}=1\)

\(d x=\frac{d t}{\sec ^{2} x}\)

\(y e^{t}=\int t \cdot \sec ^{2} x\ e^{t} \cdot d t / \sec ^{2} x\)

\(y e^{t}=\int t \cdot e^{t} \cdot d t\)

\(y e^{t}=t \int e^{t} \cdot d t-\int\left\{\frac{d}{d t}(t) \int e^{t} \cdot d t\right\} d t\)

\(y e^{t}=t e^{t}-\int e^{t} \cdot d t\)

\(y e^{t}=t e^{t}-e^{t}+C\)

Now, dividing by e^t on both sides, we get:

\(y=\tan x-1+C e^{-\tan x}\)

Therefore the solution is: \(y=\tan x-1+C e^{-\tan x}\)

Ans. Given, the differential equation = \(x d y-\left(y+2 x^{2}\right) \cdot d x=0\)

This equation can now be simplified to represent the following linear differential equation.

\(\frac{d y}{d x}=\frac{y}{x}=2 x\)

Now, comparing this with the differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = \(-\frac{1}{x}\) and Q = 2x .

Hence, we got the Integration factor as I.F which is = \(=e^{\int -\frac{1}{x}.d x}\)

\(=e^{-\log x}\)

\(=\frac{1}{x}\)

Further, the solution of the given differential equation will be as follows:

\(y(I . F)=\int Q(I . F) d x+C\)

\(y \cdot \frac{1}{x}=\int 2 x \cdot \frac{1}{y} \cdot d x+C\)

\(\frac{y}{x}=\int 2 \cdot d x+C\)

\(\frac{y}{x}=2 x+C\)

\(y=2 x^{2}+x C\)

Therefore, the solution is: \(y=2 x^{2}+x C\)

Ans. Given the slope of the tangent = dy/dx

Sum of coordinates = x + y

Therefore, \(\frac{dy}{dx}\) = x + y

\(\frac{dy}{dx}\) - 1y = x

Now, comparing this with the differential equation form: \(\frac{dy}{dx}\) + Py = Q , we have the values of P = -1 and Q = x

Hence, we got the Integration factor as I.F which is = \(e^{\int p.d x}\)

\(e^{\int -1.d x}\)

\(e^{-x}\)

Further, the solution of the given differential equation will be as follows:

\(y(I . F)=\int Q(I . F) d x+C\)

\(y e^{-x}=\int x e^{-x} d x+C\)

\(y e^{-x}=x \int e^{-x} d x-\left\{\int \frac{d}{d x}(x) \int e^{-x} d x\right\} d x+C\)

\(y e^{-x}=-x e^{-x} \int-e^{-x} d x+C\)

\(y e^{-x}=-x e^{-x}+\int\left(-e^{-x}\right)+C\)

\(y e^{-x}=-x e^{-x}-e^{-x}+C\)

Now, dividing by e^-x on both sides, we get:

\(y=-x-1+Ce^x\)

Again, for origin (0, 0) as given in the question, we get:

\(0=0-1+Ce^0\)

\(C=1\)

Therefore, \(y=-x-1+e^x\) [from \(y=-x-1+Ce^x\)]

The solution is: \(x+y+1= e^x\)