Question

Solve the following differential equations: (i) \(y y^{\prime}+1=(x-1) e^{-y^{2} / 2}\) (ii) \(y^{\prime}+x \sin 2 y=2 x e^{-x^{2}} \cos ^{2} y\) (iii) \(y y^{\prime} \sin x=\cos x\left(\sin x-y^{2}\right)\) (iv) \(y^{\prime}=\frac{y^{2}-x}{2 y(x+1)}\)

Short answer

Answer: The general solution for the third first-order non-homogeneous differential equation is \(\ln y = -y^2 \cot x + \ln(\sin x) + C\).

Step-by-step solution

Arrange the equation in explicit form

The given equation is: \(yy'+1=(x-1)e^{-y^2/2}\) Rearrange to separate and group variables: \(yy'=(x-1)e^{-y^2/2}-1\)

Integrate both sides

Integrate both sides with respect to x: \(\int y dy = \int (x-1)e^{-y^2/2}-1 dx\) Let's solve the integrals separately. On the left side, we have: \(\int y dy = \frac{1}{2}y^2 + C_1\) On the right side, integrate using substitution method. Let: \(u=-\frac{y^2}{2}\), then \(du = -yd\hspace{0.5mm}y\) We get: \(\int (x-1)e^{-y^2/2}-1 dx = \int (x-1)e^u -1 dx\)

Integrate using substitution

Now integrate the modified equation: \(\int (x-1)e^u -1 dx = xe^u - e^u - x + C_2\) Replace u with the initial value -y^2/2: \(xe^{-y^2/2} - e^{-y^2/2} - x + C_2 = \frac{1}{2}y^2 + C_1\) This is the general solution for the first-order homogeneous differential equation. #(ii): First-order non-homogeneous differential equation#

Identify the given equation

The given equation is: \(y'+x\sin 2y = 2xe^{-x^2}\cos^2 y\) This is a first-order non-homogeneous differential equation. To solve this equation, we would try to convert it into a first-order linear differential equation form.

Divide by \(\cos^2 y\)

Divide the entire equation by \(\cos^2 y\): $ \frac{y'}{\cos^2 y} + x\tan y\sec y = 2xe^{-x^2} $ Now, let \(z = \tan y\), so \(dz = \sec^2 y dy\) Thus, the given equation is transformed into: $ \frac{z'}{\cos^2 y} + xz = 2xe^{-x^2} $

Integrate both sides

Integrate both sides with respect to x: \(\int \frac{z'}{\cos^2 y} dx + \int xz dx= \int 2xe^{-x^2} dx\) Let's solve the integrals separately. On the left side, we must use integration by parts for the second integral: \(u = x\), \(dv = z dx\) \(du = dx\), \(v = \int z dx\) Integrating both sides, we have: $ \frac{z^2}{2} - \int z dx + xv - \int v dx = -e^{-x^2} + C $ Substitute back the value of z: $ \frac{\tan^2 y}{2} - \int \tan y dy + x\int \tan y dy - \int (\int \tan y dy) dy = -e^{-x^2} + C $ This is the general solution for the second first-order non-homogeneous differential equation. #(iii): First-order non-homogeneous differential equation#

Identify and rearrange the given equation

The given equation is: \(yy' \sin x = \cos x (\sin x - y^2)\) Rearrange to have the form: \(y' = \frac{\cos x (\sin x - y^2)}{y\sin x}\) Now we integrate both sides.

Integrate both sides

Integrate both sides with respect to x: \(\int \frac{1}{y} dy = \int \frac{\cos x (\sin x - y^2)}{\sin x} dx\) Solve the integrals explicitly, and we have: $ \ln y = -y^2 \cot x + \ln(\sin x) + C $ This is the general solution for the third first-order non-homogeneous differential equation. #(iv): First-order non-homogeneous differential equation#

Identify and rearrange the given equation

The given equation is: \(y' = \frac{y^2 - x}{2 y(x+1)}\) Now we integrate both sides.

Integrate both sides

Integrate both sides with respect to x: \(\int dy = \int \frac{y^2 - x}{2 y(x+1)} dx\) Solve the integrals explicitly, and we have: $ y = \frac{x^2 + 2xy + 2y^2 - 2C}{(x+1)^2} $ This is the general solution for the fourth first-order non-homogeneous differential equation.